Thermodynamics Made Simple for Energy Engineers: & Engineers in Other Disciplines [2 ed.] 8770223491, 9788770223492

Table of contents :
Cover
Half Title
Series Page
Title Page
Copyright Page
Dedication
Table of Contents
Preface
List of Figures
List of Tables
Chapter 1: Introduction to Energy, Heat and Thermodynamics
1.1: Introduction
1.2: Energy
1.3: Root Concepts and Terms that Contribute toward the Production or Transformation of Energy
1.3.1: Force and mass
1.3.2: Density and weight density
1.3.3: Specific volume
1.3.4: Pressure
1.3.5: Temperature
1.3.6: Absolute temperature
1.3.7: Law of conservation of energy
1.4: Forms of Energy in Mechanical and Thermodynamic Systems
1.4.1: Potential energy
1.4.2: Kinetic energy
1.4.3: Energy stored in a spring
1.4.4: Pressure energy
1.4.5: Heat and internal energy of a system
1.4.6: Unit conversions3: associated with heat energy
1.4.7: Molar internal energy
1.5: Case Study 1.1: Energy and Energy Unit Conversion
1.5.1: Work
1.5.2: Work in a mechanical system
1.5.3: Mathematical equations for work
1.5.4: Work performed in a thermodynamic system
1.5.5: Specific heat
1.6: Case Study 1.2: Energy Conservation, Energy Conversion and Thermodynamics
Chapter 1—Self-Assessment Problem and Question
Chapter 2: Thermodynamics and Power
2.1: Introduction
2.2: Power and Efficiency
2.2.1: Power
2.2.2: Units for power
2.2.3: Common power conversion factors in the SI system
2.2.4: Units for energy
2.2.5: SI or Metric unit system
2.2.6: Common energy conversion factors
2.2.7: Efficiency
2.2.8: Power – steam, mechanical and electrical
2.3: Case Study 2.1. Steam to Electricity Conversion
Chapter 2—Self-Assessment Problems and Questions
Chapter 3: Study of Enthalpy and Entrophy
3.1: Introduction
3.1.1: Enthalpy
3.1.2: Entropy
3.2: Case Study 3.1: Entropy Analysis
Chapter 3—Self-Assessment Problems and Questions
Chapter 4: Understanding Mollier Diagram
4.1: Introduction
4.2: Application of Mollier Diagram
4.2.1: Enthalpy determination
4.2.2: Entropy determination
Chapter 4—Self-Assessment Problems and Questions
Chapter 5: Saturated and Superhcteated Steam Tables
5.1: Introduction
5.2: Saturated Steam Tables
5.3: Superheated Steam Tables
5.4: Single and Double Interpolation of Steam Table Data
5.5: Quality of Steam Consideration in Thermodynamic Calculations
Chapter 5—Self-Assessment Problems and Questions
Chapter 6: Phases of Water and Associated Thermodynamics
6.1: Phases of Substance
6.1.1: Solid
6.1.2: Liquid
6.1.3: Gas
6.1.4: Sensible heat
6.1.5: Latent heat
6.1.6: Saturation temperature
6.1.7: Saturation pressure
6.1.8: Subcooled liquid
6.1.9: Saturated liquid
6.1.10: Saturated vapor
6.1.11: Liquid-vapor phase
6.1.12: Superheated vapor
6.2: Phase Transformation of Water at Constant Pressure
6.3: Types of Phase Transformation
6.3.1: Ideal gas
6.3.2: Real gas
6.3.3: Critical point
6.3.4: Critical properties
6.3.5: Triple point
6.3.6: Comparison - triple point vs. critical point
Chapter 6—Self-Assessment Problems and Questions
Chapter 7: Laws of Thermodynamics
7.1: Introduction
7.2: Major Categories of Thermodynamic Systems
7.2.1: Open thermodynamic systems
7.2.2: Closed thermodynamic systems
7.2.3: Isolated thermodynamic systems
7.3: Laws of Thermodynamics
7.3.1: First law of thermodynamics systems
7.3.2: Second law of thermodynamics
7.4: Case Study 7.1: SI Units
7.5: Case Study Solution Strategy
7.6: Case Study 7.1: US/Imperial Units, With Illustration of Interpolation Method
7.7: Case Study Solution Strategy
7.7.1: Single and double interpolation of steam table data for enthalpy determination
Chapter 7—Self-Assessment Problems and Questions
Chapter 8: Thermodynamic Processes
8.1: Introduction
8.2: Thermodynamic Processes
8.2.1: Adiabatic process
8.3: Adiabatic Process Example I – Throttling Process in a Refrigeration System
8.4: Adiabatic Process Example II – Compressor Segment of a Refrigeration System
8.4.1: Isenthalpic or isoenthalpic process
8.4.2: Isenthalpic process example – Throttling process in a refrigeration system
8.4.3: Constant pressure or isobaric process
8.4.4: Isobaric process example I: Evaporation stage of a refrigeration cycle
8.4.5: Isobaric process example II: Isobaric segments of an ideal cycle heat engine
8.4.6: Constant temperature or isothermal process
8.4.7: Isothermal process example I: Steam generation process
8.4.8: Constant volume process
8.4.9: Constant volume process example I: Superheated steam generation in a “rigid” constant volume boiler
8.4.10: Constant volume process example II: Ideal heat engine
8.4.11: Isentropic or constant entropy process
8.4.12: Isentropic process example I: Ideal heat engine – carnot cycle
8.4.13: Throttling process and inversion point
8.4.14: Thermodynamic equilibrium
8.4.15: Quasistatic or quasiequilibrium process
8.4.16: Polytropic process
8.4.17: Reversible process
8.4.18: Irreversible process
8.4.19: Ideal heat engine, ideal heat engine cycle and energy flow
8.4.20: Reaction turbine
8.4.21: Impulse turbine
8.4.22: Process flow in a rankine cycle with superheat
8.4.23: Rankine cycle equations
8.5: Case Study 8.1: Rankine Engine
8.5.1: Carnot cycle
8.5.2: Carnot cycle equations
8.5.3: Comparison between rankine and carnot cycles
8.5.4: Other major types of cycles
8.5.5: Cogeneration
8.5.6: Combined cycle
Chapter 8—Self-Assessment Problems and Questions
Chapter 9: Gas Dynamics
9.1: Introduction
9.2: Steady Flow Energy Equation
9.3: Case Study 9.1
9.3.1: SI unit system
9.3.2: US unit system
9.4: Isentropic Flow
9.4.1: Critical point (gas dynamics)
9.4.2: Shock waves
Chapter 9—Self-Assessment Problems and Questions
Chapter 10: Psychrometry and Psychrometric Analysis
10.1: Introduction
10.2: The Psychrometric Chart
10.2.1: Dry-bulb temperature (DB)
10.2.2: Wet-bulb temperature (WB)
10.2.3: Dew-point temperature (DP)
10.2.4: Relative humidity (RH)
10.2.5: Humidity ratio
10.2.6: Specific enthalpy
10.3: Method for Reading the Psychrometric Chart
10.4: Psychrometric Transition Process
10.5: Case Study 10.1: Psychrometrics – SI Unit System
10.6: Case Study 10.2: Psychrometrics – US Unit System
Chapter 10—Self-Assessment Problems and Questions
Chapter 11: Refrigeration Cycles and HVAC Systems
11.1: Introduction
11.2: Types of Air Conditioning Systems
11.2.1: Refrigeration system compressors
11.2.2: Refrigeration system condenser
11.2.3: Refrigerants
11.2.4: Expansion valve
11.2.5: Cooling capacity of refrigeration systems
11.2.6: Refrigeration system capacity quantification in A/C tons
11.2.7: Basic refrigeration cycle
11.3: Refrigerant Compression
11.3.1: Wet vapor compression process
11.3.2: Refrigerant vapor quality ratio
11.3.3: Dry vapor compression process
11.3.4: Coefficient of performance, or COP, in refrigeration systems – refrigerator example
11.4: SEER, Seasonal Energy Efficiency Ratio
11.5: Case Study 11.1: Refrigeration Cycle
11.6: Direct Digital Control of HVAC Systems
11.6.1: Digital or discrete inputs
11.6.2: Digital or discrete outputs
11.6.3: Analog inputs
11.6.4: Analog outputs
Chapter 11—Self-Assessment Problems and Questions
Appendices
Appendix A
Chapter 1—Self-Assessment Problems and Questions
Chapter 2—Self-Assessment Problems and Questions
Chapter 3—Self-Assessment Problems and Questions
Chapter 4—Self-Assessment Problems and Questions
Chapter 5—Self-Assessment Problems and Questions
Chapter 6—Self-Assessment Problems and Questions
Chapter 7—Self-Assessment Problems and Questions
Chapter 8—Self-Assessment Problems and Questions
Chapter 9—Self-Assessment Problems and Questions
Chapter 10—Self-Assessment Problems and Questions
Chapter 11—Self-Assessment Problems and Questions
Appendix B
Steam Tables
Appendix C
Common Units and Unit Conversion Factors
Appendix D
Common Symbols
Index
About the Author

Citation preview

Thermodynamics Made Simple for Energy Engineers & Engineers in Other Disciplines

Second Edition

RIVER PUBLISHERS SERIES IN ENERGY ENGINEERING AND SYSTEMS Series Editors BOBBY RAUF Semtrain, LLC, USA The "River Publishers Series in Energy Engineering and Systems" is a series of comprehensive academic and professional books focussing on the theory and applications behind various energy-related technologies and control systems. The series features handbooks for related technology, as well as manuals on the fundamentals and theoretical aspects of energy engineering projects. The main aim of this series is to be a reference for academics, researchers, managers, engineers, and other professionals in related matters with energy engineering and control systems. Topics covered in the series include, but are not limited to:

• Energy engineering; • Machinery; • Testing; • HVAC; • Electricity and electronics; • Water systems; • Industrial control systems; • Web based systems; • Automation; • Lighting controls. For a list of other books in this series, visit www.riverpublishers.com

Thermodynamics Made Simple for Energy Engineers & Engineers in Other Disciplines Second Edition

S. Bobby Rauf Sem-Train, USA

River Publishers

Published 2023 by River Publishers River Publishers Alsbjergvej 10, 9260 Gistrup, Denmark www.riverpublishers.com Distributed exclusively by Routledge

605 Third Avenue, New York, NY 10017, USA 4 Park Square, Milton Park, Abingdon, Oxon OX14 4RN

Thermodynamics Made Simple for Energy Engineers & Engineers in Other Disciplines – Second Edition / S. Bobby Rauf. ©2023 River Publishers. All rights reserved. No part of this publication may be reproduced, stored in a retrieval systems, or transmitted in any form or by any means, mechanical, photocopying, recording or otherwise, without prior written permission of the publishers. Routledge is an imprint of the Taylor & Francis Group, an informa business ISBN 978-87-7022-349-2 (print) ISBN 978-87-7022-959-3 (paperback) ISBN 978-10-0092-368-1 (online) ISBN 978-1-003-42761-2 (ebook master) While every effort is made to provide dependable information, the publisher, authors, and editors cannot be held responsible for any errors or omissions.

This book is dedicated to my uncle Dr. Stanley Peters, without whom, this wouldn’t be!

Contents

Preface

xv

List of Figures

xix

List of Tables

xxiii

1 Introduction to Energy, Heat and Thermodynamics 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 1.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Root Concepts and Terms that Contribute toward the Production or Transformation of Energy . . . . . . . 1.3.1 Force and mass . . . . . . . . . . . . . . . . 1.3.2 Density and weight density . . . . . . . . . . 1.3.3 Specific volume . . . . . . . . . . . . . . . . 1.3.4 Pressure . . . . . . . . . . . . . . . . . . . . 1.3.5 Temperature . . . . . . . . . . . . . . . . . . 1.3.6 Absolute temperature . . . . . . . . . . . . . 1.3.7 Law of conservation of energy . . . . . . . . 1.4 Forms of Energy in Mechanical and Thermodynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Potential energy . . . . . . . . . . . . . . . . 1.4.2 Kinetic energy . . . . . . . . . . . . . . . . . 1.4.3 Energy stored in a spring . . . . . . . . . . . 1.4.4 Pressure energy . . . . . . . . . . . . . . . . 1.4.5 Heat and internal energy of a system . . . . . 1.4.6 Unit conversions3 associated with heat energy 1.4.7 Molar internal energy . . . . . . . . . . . . . 1.5 Case Study 1.1: Energy and Energy Unit Conversion 1.5.1 Work . . . . . . . . . . . . . . . . . . . . . 1.5.2 Work in a mechanical system . . . . . . . . . 1.5.3 Mathematical equations for work . . . . . . .

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viii  Contents 1.5.4 Work performed in a thermodynamic system . . . . 1.5.5 Specific heat . . . . . . . . . . . . . . . . . . . . 1.6 Case Study 1.2: Energy Conservation, Energy Conversion and Thermodynamics . . . . . . . . . . . . . . . . . . . . Chapter 1—Self-Assessment Problem and Question . . . . . . . 2 Thermodynamics and Power 2.1 Introduction . . . . . . . . . . . . . . . . . . . . 2.2 Power and Efficiency . . . . . . . . . . . . . . . 2.2.1 Power . . . . . . . . . . . . . . . . . . . 2.2.2 Units for power . . . . . . . . . . . . . . 2.2.3 Common power conversion factors in the SI system . . . . . . . . . . . . . . . 2.2.4 Units for energy . . . . . . . . . . . . . . 2.2.5 SI or Metric unit system . . . . . . . . . . 2.2.6 Common energy conversion factors . . . . 2.2.7 Efficiency . . . . . . . . . . . . . . . . . 2.2.8 Power – steam, mechanical and electrical . 2.3 Case Study 2.1. Steam to Electricity Conversion . Chapter 2—Self-Assessment Problems and Questions .

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3 Study of Enthalpy and Entrophy 3.1 Introduction . . . . . . . . . . . . . . . . . . . 3.1.1 Enthalpy . . . . . . . . . . . . . . . . . 3.1.2 Entropy . . . . . . . . . . . . . . . . . 3.2 Case Study 3.1: Entropy Analysis . . . . . . . . Chapter 3—Self-Assessment Problems and Questions

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4 Understanding Mollier Diagram 4.1 Introduction . . . . . . . . . . . . . . . . . . . 4.2 Application of Mollier Diagram . . . . . . . . . 4.2.1 Enthalpy determination . . . . . . . . . 4.2.2 Entropy determination . . . . . . . . . . Chapter 4—Self-Assessment Problems and Questions

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5 Saturated and Superhcteated Steam Tables 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 5.2 Saturated Steam Tables . . . . . . . . . . . . . . . . 5.3 Superheated Steam Tables . . . . . . . . . . . . . . . 5.4 Single and Double Interpolation of Steam Table Data

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Contents  ix

5.5

Quality of Steam Consideration in Thermodynamic Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5—Self-Assessment Problems and Questions . . . . . . .

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6 Phases of Water and Associated Thermodynamics 6.1 Phases of Substance . . . . . . . . . . . . . . . . . . 6.1.1 Solid . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Liquid . . . . . . . . . . . . . . . . . . . . . 6.1.3 Gas . . . . . . . . . . . . . . . . . . . . . . . 6.1.4 Sensible heat . . . . . . . . . . . . . . . . . . 6.1.5 Latent heat . . . . . . . . . . . . . . . . . . . 6.1.6 Saturation temperature . . . . . . . . . . . . 6.1.7 Saturation pressure . . . . . . . . . . . . . . 6.1.8 Subcooled liquid . . . . . . . . . . . . . . . . 6.1.9 Saturated liquid . . . . . . . . . . . . . . . . 6.1.10 Saturated vapor . . . . . . . . . . . . . . . . 6.1.11 Liquid-vapor phase . . . . . . . . . . . . . . 6.1.12 Superheated vapor . . . . . . . . . . . . . . . 6.2 Phase Transformation of Water at Constant Pressure . 6.3 Types of Phase Transformation . . . . . . . . . . . . 6.3.1 Ideal gas . . . . . . . . . . . . . . . . . . . . 6.3.2 Real gas . . . . . . . . . . . . . . . . . . . . 6.3.3 Critical point . . . . . . . . . . . . . . . . . . 6.3.4 Critical properties . . . . . . . . . . . . . . . 6.3.5 Triple point . . . . . . . . . . . . . . . . . . 6.3.6 Comparison - triple point vs. critical point . . Chapter 6—Self-Assessment Problems and Questions . . .

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7 Laws of Thermodynamics 7.1 Introduction . . . . . . . . . . . . . . . . . . 7.2 Major Categories of Thermodynamic Systems 7.2.1 Open thermodynamic systems. . . . . 7.2.2 Closed thermodynamic systems . . . . 7.2.3 Isolated thermodynamic systems . . . 7.3 Laws of Thermodynamics . . . . . . . . . . . 7.3.1 First law of thermodynamics systems . 7.3.2 Second law of thermodynamics . . . . 7.4 Case Study 7.1: SI Units. . . . . . . . . . . . 7.5 Case Study Solution Strategy . . . . . . . . .

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Case Study 7.1: US/Imperial Units, With Illustration of Interpolation Method . . . . . . . . . . . . . . . . . . . . . 7.7 Case Study Solution Strategy . . . . . . . . . . . . . . . . . 7.7.1 Single and double interpolation of steam table data for enthalpy determination . . . . . . . . . . . . . . . . Chapter 7—Self-Assessment Problems and Questions . . . . . . .

8 Thermodynamic Processes 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Thermodynamic Processes . . . . . . . . . . . . . . . . . . 8.2.1 Adiabatic process . . . . . . . . . . . . . . . . . . . 8.3 Adiabatic Process Example I – Throttling Process in a Refrigeration System . . . . . . . . . . . . . . . . . . . . . 8.4 Adiabatic Process Example II – Compressor Segment of a Refrigeration System . . . . . . . . . . . . . . . . . . . . . 8.4.1 Isenthalpic or isoenthalpic process . . . . . . . . . . 8.4.2 Isenthalpic process example – Throttling process in a refrigeration system . . . . . . . . . . . . . . . . . . 8.4.3 Constant pressure or isobaric process . . . . . . . . . 8.4.4 Isobaric process example I: Evaporation stage of a refrigeration cycle. . . . . . . . . . . . . . . . . 8.4.5 Isobaric process example II: Isobaric segments of an ideal cycle heat engine . . . . . . . . . . . . . 8.4.6 Constant temperature or isothermal process . . . . . 8.4.7 Isothermal process example I: Steam generation process . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.8 Constant volume process . . . . . . . . . . . . . . . 8.4.9 Constant volume process example I: Superheated steam generation in a “rigid” constant volume boiler . . . . . . . . . . . . . . . . . . . . . 8.4.10 Constant volume process example II: Ideal heat engine . . . . . . . . . . . . . . . . . . . . 8.4.11 Isentropic or constant entropy process . . . . . . . . 8.4.12 Isentropic process example I: Ideal heat engine – carnot cycle . . . . . . . . . . . . . . . . . . . . . . 8.4.13 Throttling process and inversion point . . . . . . . . 8.4.14 Thermodynamic equilibrium . . . . . . . . . . . . . 8.4.15 Quasistatic or quasiequilibrium process . . . . . . . . 8.4.16 Polytropic process . . . . . . . . . . . . . . . . . . .

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Contents  xi

8.4.17 Reversible process . . . . . . . . . . . . . . . . 8.4.18 Irreversible process . . . . . . . . . . . . . . . 8.4.19 Ideal heat engine, ideal heat engine cycle and energy flow . . . . . . . . . . . . . . . . . . . 8.4.20 Reaction turbine . . . . . . . . . . . . . . . . . 8.4.21 Impulse turbine . . . . . . . . . . . . . . . . . 8.4.22 Process flow in a rankine cycle with superheat . 8.4.23 Rankine cycle equations . . . . . . . . . . . . . 8.5 Case Study 8.1: Rankine Engine . . . . . . . . . . . . 8.5.1 Carnot cycle . . . . . . . . . . . . . . . . . . . 8.5.2 Carnot cycle equations . . . . . . . . . . . . . 8.5.3 Comparison between rankine and carnot cycles 8.5.4 Other major types of cycles . . . . . . . . . . . 8.5.5 Cogeneration . . . . . . . . . . . . . . . . . . 8.5.6 Combined cycle . . . . . . . . . . . . . . . . . Chapter 8—Self-Assessment Problems and Questions . . . .

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9 Gas Dynamics 9.1 Introduction . . . . . . . . . . . . . . . . . . . 9.2 Steady Flow Energy Equation . . . . . . . . . . 9.3 Case Study 9.1 . . . . . . . . . . . . . . . . . . 9.3.1 SI unit system . . . . . . . . . . . . . . 9.3.2 US unit system. . . . . . . . . . . . . . 9.4 Isentropic Flow . . . . . . . . . . . . . . . . . 9.4.1 Critical point (gas dynamics) . . . . . . 9.4.2 Shock waves . . . . . . . . . . . . . . . Chapter 9—Self-Assessment Problems and Questions

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10 Psychrometry and Psychrometric Analysis 10.1 Introduction . . . . . . . . . . . . . . . . . . 10.2 The Psychrometric Chart . . . . . . . . . . . 10.2.1 Dry-bulb temperature (DB) . . . . . . 10.2.2 Wet-bulb temperature (WB) . . . . . . 10.2.3 Dew-point temperature (DP) . . . . . 10.2.4 Relative humidity (RH) . . . . . . . . 10.2.5 Humidity ratio . . . . . . . . . . . . . 10.2.6 Specific enthalpy . . . . . . . . . . . 10.3 Method for Reading the Psychrometric Chart . 10.4 Psychrometric Transition Process . . . . . .

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xii  Contents 10.5 Case Study 10.1: Psychrometrics – SI Unit System. . . . . . 10.6 Case Study 10.2: Psychrometrics – US Unit System . . . . . Chapter 10—Self-Assessment Problems and Questions . . . . . . 11 Refrigeration Cycles and HVAC Systems 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Types of Air Conditioning Systems . . . . . . . . . . . . . 11.2.1 Refrigeration system compressors. . . . . . . . . . 11.2.2 Refrigeration system condenser . . . . . . . . . . . 11.2.3 Refrigerants . . . . . . . . . . . . . . . . . . . . . 11.2.4 Expansion valve . . . . . . . . . . . . . . . . . . . 11.2.5 Cooling capacity of refrigeration systems . . . . . . 11.2.6 Refrigeration system capacity quantification in A/C tons . . . . . . . . . . . . . . . . . . . . . . 11.2.7 Basic refrigeration cycle . . . . . . . . . . . . . . . 11.3 Refrigerant Compression . . . . . . . . . . . . . . . . . . 11.3.1 Wet vapor compression process . . . . . . . . . . . 11.3.2 Refrigerant vapor quality ratio . . . . . . . . . . . 11.3.3 Dry vapor compression process . . . . . . . . . . . 11.3.4 Coefficient of performance, or COP, in refrigeration systems – refrigerator example . . . . . . . . . . . 11.4 SEER, Seasonal Energy Efficiency Ratio . . . . . . . . . 11.5 Case Study 11.1: Refrigeration Cycle . . . . . . . . . . . . 11.6 Direct Digital Control of HVAC Systems . . . . . . . . . 11.6.1 Digital or discrete inputs . . . . . . . . . . . . . . 11.6.2 Digital or discrete outputs . . . . . . . . . . . . . . 11.6.3 Analog inputs . . . . . . . . . . . . . . . . . . . . 11.6.4 Analog outputs . . . . . . . . . . . . . . . . . . . Chapter 11—Self-Assessment Problems and Questions . . . . .

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236 237 238 239 240 241

. . . . . . . . .

242 243 244 248 252 252 254 254 256

Appendices Appendix A Chapter 1—Self-Assessment Problems and Questions Chapter 2—Self-Assessment Problems and Questions Chapter 3—Self-Assessment Problems and Questions Chapter 4—Self-Assessment Problems and Questions Chapter 5—Self-Assessment Problems and Questions Chapter 6—Self-Assessment Problems and Questions Chapter 7—Self-Assessment Problems and Questions

212 218 223

259 . . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

259 259 260 262 265 267 273 277

Contents  xiii

Chapter 8—Self-Assessment Problems and Questions . Chapter 9—Self-Assessment Problems and Questions . Chapter 10—Self-Assessment Problems and Questions Chapter 11—Self-Assessment Problems and Questions

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

292 295 299 308

Appendix B Steam Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

315 315

Appendix C Common Units and Unit Conversion Factors . . . . . . . . . . . .

337 337

Appendix D Common Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .

341 341

Index

343

About the Author

353

Preface

Like the first edition of this text, this edition is designed to serve as a tool for building basic engineering skills in the realm of thermodynamics and HVAC. Similar to new editions of most publications, readers will find copious updates to technical data and case-studies in the spirit of continuous improvement. Some of the updates, revisions, and improvements in this edition are as follows: ●●

Improved formula formats.

●●

Expanded the discussion and explanation on differences and similarities between Mollier Diagram and Psychrometric Chart. More illustrative diagrams added.

●●

Station Zebra Case Study calibrated and enhanced to align to more realistic turbine efficiency levels.

●●

Enhanced discussion and explanation on the topic of phases of water.

●●

Current energy cost information and trend.

●●

Discussion and explanation on ionization of gas to form plasma and deionization of plasma to revert back to plain gaseous phase has been added.

●●

Better explanation of “critical Point.”

●●

Better, clearer, explanation of operation of heat engines.

●●

More diagrams added to enhance the explanation of various thermodynamics concepts.

●●

Clarification of “critical point” in gas dynamics. New illustration of shock waves in supersonic flight of aircraft.

●●

Additional explanation on the subject of psychrometry.

●●

An update on soon-to-be-banned refrigerants and their potential replacements.

●●

Updated refrigeration cycle diagram. xv

xvi  Preface If your objective as a reader is limited to the acquisition of basic knowledge in thermodynamics and psychrometrics, then the material in this book should suffice. If, however, the reader wishes to progress their knowledge and skills in thermodynamics and psychrometry to intermediate or advanced level, this book could serve as an effective stepping-stone. As the adage goes, “a picture is worth a thousand words;” this book maximizes the utilization of diagram, graphs, and flow charts to facilitate quick and effective comprehension of the concepts of thermodynamics by the reader. In this book, the study of thermodynamics concepts, principles and analysis techniques is made relatively easy for the reader by inclusion of most of the reference data, in form of excerpts, within the discussion of each case study, exercise, and self-assessment problem solutions. This is to facilitate quick study and comprehension of the material without repetitive and time-consuming search for reference data in other parts of the book. Certain thermodynamic concepts and terms are explained more than once as these concepts are reintroduced in different chapters of this text; often with a slightly different perspective and context. This approach is a deliberate attempt to make the study of some of the more abstract thermodynamics topics more fluid; affording the reader continuity and precluding the need for pausing and referring to chapters where those specific topics were first introduced. Due to the level of explanation and details included on thermodynamics concepts, principles, computational techniques and analytical methods, this book provides the opportunity for engineers and non-engineers, who are not current on the subject of thermodynamics, an opportunity to refresh and renew their thermodynamics and HVAC skills and knowledge. The solutions for end-of-the-chapter self-assessment problems are explained in just as much detail as the pertaining case studies and sample problem in the respective chapters. Since all chapters and topics begin with the introduction of important fundamental concepts and principles, this book can serve as a “brush-up” or review tool for even mechanical engineers whose current area of engineering specialty does not afford them the opportunity to keep their thermodynamics knowledge current. To clarify some of the thermodynamic concepts effectively for professionals whose engineering education focus does not include thermodynamics, analogies are drawn from non-mechanical engineering realms on certain complex topics, to facilitate comprehension of the relatively abstract thermodynamic concepts and principles.

Preface  xvii

Each chapter in this book concludes with a list of questions or problems, for self-assessment, skill building and knowledge affirmation purposes. The reader is encouraged to attempt these problems and questions. The answers and solutions, for the questions and problems, are included under Appendix A of this text. For reference and computational purposes, steam tables and Mollier (Enthalpy-Entropy) diagrams are included in Appendix B. Most engineers understand the role units play in definition and verification of the engineering concepts, principles, equations, and analytical techniques. Therefore, most thermodynamic concepts, principles and computational procedures covered in this book are punctuated with proper units. In addition, for the reader’s convenience, units for commonly used thermodynamic entities, and some conversion factors are listed under Appendix C. Most thermodynamic concepts, principles, tables, graphs, and computational procedures covered in this book are premised on US/Imperial Units as well as SI/Metric Units. Certain numerical examples, case studies or self-assessment problems in this book are premised on either the SI Unit realm or the US Customary Unit system. When the problems or numerical analysis are based on only one of the two unit systems, the given data and the final results can be transformed into the desired unit system through the use of unit conversion factors in Appendix C. Some of the Greek symbols, used in the realm of thermodynamics, are listed in Appendix D, for reference.

What readers can gain from this book: ●●

Better understanding of thermodynamics terms, concepts, principles, laws, analysis methods, solution strategies and computational techniques.

●●

Better understanding of HVAC terms, concepts, principles, laws, analysis methods, solution strategies and computational techniques.

●●

Differences and similarities between Mollier’s Diagram and Psychrometric Chart and their applications as analytical tools for performing HVAC and steam system analyses.

●●

Better understanding of gas dynamics and pertinent analytical techniques.

●●

Greater confidence in interactions with thermodynamics design engineers and thermodynamics experts.

xviii  Preface ●●

Skills and preparation necessary for succeeding in thermodynamics portion of various certification and licensure exams, i.e., CEM, FE, PE, and many other trade certification tests.

●●

A better understanding of the thermodynamics component of heat related energy projects.

●●

A compact and simplified thermodynamics desk reference that includes compact saturated steam tables and superheated steam tables.

List of Figures

Figure 1.1 Figure 1.2

Uranium fission reaction. . . . . . . . . . . . . . . Illustration of mechanical work, in a system with friction. . . . . . . . . . . . . . . . . . . . . . . . Figure 1.3 Case Study 1.2. . . . . . . . . . . . . . . . . . . . Figure 2.1 Steam to Wire Power Flow – Steam Power Generation System. . . . . . . . . . . . . . . . . . Figure 3.1 Case Study 3.1, Entropy. . . . . . . . . . . . . . . Figure 4.1 Mollier diagram, SI/Metric units. . . . . . . . . . . Figure 4.2 Mollier diagram, US customary units. . . . . . . . Figure 4.3 Mollier diagram transformation to psychrometric chart. . . . . . . . . . . . . . . . . . . . . . . . . Figure 4.4 Mollier diagram, SI/Metric units. . . . . . . . . . . Figure 5.1 Mollier diagram, SI/Metric units. . . . . . . . . . . Figure 6.1 Phase transformation of water at constant pressure. Figure 6.2 Phase transformation of water at constant pressure, depicted in graphical form. . . . . . . . . . . . . . Figure 6.3 Phase transformation of water. . . . . . . . . . . . Figure 6.4 Comparison of triple point vs. critical point, a graphical perspective. . . . . . . . . . . . . . . . . Figure 7.1 Open thermodynamic system with a turbine. . . . . Figure 7.2 Thermodynamic process flow diagram, case study 7.1. . . . . . . . . . . . . . . . . . . . . Figure 7.2(A) natural gas pricing at the time of publication of this text. . . . . . . . . . . . . . . . . . . . . . . . Figure 7.3 Case study 7.1, mass flow rate analysis. . . . . . . Figure 7.4 Case study 7.1 stage 1 sensible heat calculation. . . Figure 7.5 Case study 7.1 stage 2 latent heat calculation. . . . Figure 7.6 Case study 7.1 stage 3 sensible heat calculation. . . Figure 7.7 Case study 7.1 stage 4 latent heat calculation. . . . Figure 7.8 Case study 7.1 stage 5 sensible heat calculation. . . Figure 7.9 Thermodynamic process flow diagram, Case study 7.1. . . . . . . . . . . . . . . . . . . . xix

.

11

. .

13 18

. . . .

35 43 48 49

. . . .

50 52 72 83

. .

84 85

. .

88 95

.

101

. . . . . . .

102 104 107 109 110 112 113

.

120

xx  List of Figures Figure 7.10 Figure 7.11 Figure 7.12 Figure 7.13 Figure 7.14 Figure 7.15 Figure 8.1 Figure 8.2 Figure 8.3 Figure 8.4 Figure 8.5 Figure 8.6 Figure 8.7 Figure 8.8 Figure 8.9 Figure 8.10 Figure 8.11 Figure 8.12 Figure 8.13 Figure 8.14 Figure 8.15 Figure 8.16 Figure 8.17 Figure 8.18 Figure 8.19 Figure 8.20 Figure 8.21 Figure 8.22

Case study 7.1, Mass flow rate analysis, US units. . . Case study 7.1, US units, stage 1 sensible heat calculation. . . . . . . . . . . . . . . . . . . . . . . Case study 7.1, US units, stage 2 latent heat calculation. . . . . . . . . . . . . . . . . . . . . . . Case study 7.1, US units, stage 3 sensible heat calculation. . . . . . . . . . . . . . . . . . . . . . . Case study 7.1 stage 4 latent heat calculation. . . . . Case study 7.1 stage 5 sensible heat calculation. . . . Work performed in a real gas, adiabatic, thermodynamic process. . . . . . . . . . . . . . . . Work performed in an ideal adiabatic thermodynamic process. . . . . . . . . . . . . . . . Thermodynamic processes in a typical refrigeration cycle. . . . . . . . . . . . . . . . . . . Heat absorbed in an isobaric thermodynamic process. Isobaric process in an ideal heat engine cycle. . . . . Isothermal process in a heat engine heat cycle. . . . . Fire-tube boiler design and operation concept diagram. . . . . . . . . . . . . . . . . . . . . . . . . Isentropic processes in a carnot cycle, an ideal cycle heat engine. . . . . . . . . . . . . . . . . . . . Heat engine energy flow diagram. . . . . . . . . . . Heat engine process flow diagram. . . . . . . . . . . Concentrated solar power tower – Solucor/Estela project. . . . . . . . . . . . . . . . . . . . . . . . . Geothermal electrical power generation process. . . . Heat cycle in a heat engine without superheat. . . . . Heat cycle in a heat engine without superheat. . . . . Heat cycle in a rankine engine with superheat. . . . . Reaction turbine history and operating principle. . . Impulse turbine design principle. . . . . . . . . . . . Couple in an impulse turbine. . . . . . . . . . . . . . Heat cycle in a rankine engine with superheat. . . . . Heat cycle in a rankine engine with superheat, p versus n. . . . . . . . . . . . . . . . . . . . . . . . Heat Cycle in a Rankine Engine with Superheat, T versus s. . . . . . . . . . . . . . . . . . . . . . . . Heat cycle in a rankine engine with superheat, h versus n. . . . . . . . . . . . . . . . . . . . . . . .

125 131 132 133 135 136 147 148 150 151 153 155 158 159 162 162 164 165 165 166 166 170 170 170 172 172 173 174

List of Figures  xxi

Figure 8.23 Figure 8.24 Figure 8.25 Figure 8.26 Figure 8.27 Figure 8.28 Figure 8.29 Figure 8.30 Figure 8.31 Figure 8.32 Figure 8.33 Figure 9.1 Figure 9.2 Figure 9.3 Figure 10.1 Figure 10.2 Figure 10.3 Figure 10.4 Figure 10.5 Figure 10.6 Figure 10.7 Figure 11.1 Figure 11.2 Figure 11.3 Figure 11.4 Figure 11.5 Figure 11.6 Figure 11.7 Figure 11.8

Heat cycle in a rankine engine with superheat, h versus n. Case study 8.1. . . . . . . . . . . . . . . Heat cycle in a rankine engine with superheat, h versus n; Case study 8.1. . . . . . . . . . . . . . . Heat cycle in a carnot engine, p versus n.. . . . . . . Carnot heat cycle, T versus s. . . . . . . . . . . . . . Carnot heat cycle, h versus s. . . . . . . . . . . . . . Alternative representation of carnot heat Cycle, p versus u. . . . . . . . . . . . . . . . . . . . . . . . Rankinee and carnot cycle comparison, temperature versus entropy. . . . . . . . . . . . . . . Working principle of a rankine cycle engine with superheat and reheat. . . . . . . . . . . . . . . . . . Working principle of a binary cycle system. . . . . . Working principle of a combined cycle system. . . . Heat cycle in a rankine engine with superheat, h versus s. . . . . . . . . . . . . . . . . . . . . . . . High velocity flow. . . . . . . . . . . . . . . . . . . High velocity flow, case study 9.1. . . . . . . . . . . Orthogonal propagation of shock waves. . . . . . . . Psychrometric chart – Copyright and Courtesy AAON. . . . . . . . . . . . . . . . . . . . . . . . . Psychrometric chart reading guide. . . . . . . . . . . Psychrometric processes. . . . . . . . . . . . . . . . Psychrometric chart – Case study 10.1, SI unit system. . . . . . . . . . . . . . . . . . . . . . . . . Psychrometric chart – Case study 10.2. . . . . . . . . Psychrometric chart – Problem 1.. . . . . . . . . . . Psychrometric chart – Problem 2.. . . . . . . . . . . Large refrigeration system chiller vs. a refrigerator compressor. . . . . . . . . . . . . . . . . . . . . . . Open air washer system architecture. . . . . . . . . . Forced air type condenser cooling tower for refrigeration system. . . . . . . . . . . . . . . . . . Refrigeration system thermal expansion valve. . . . . Refrigeration cycle process flow diagram. . . . . . . Refrigeration cycle pressure-enthalpy graph. . . . . . Wet vapor compression cycle in refrigeration systems.. . Dry vapor compression cycle in refrigeration systems. . . . . . . . . . . . . . . . . . . . . . . . .

179 180 183 184 185 185 188 189 190 191 192 196 198 200 206 207 211 214 221 224 225 228 230 231 235 237 239 240 242

xxii  List of Figures Figure 11.9 Figure 11.10 Figure 11.11 Figure 11.12 Figure A-3.1 Figure A-4.1 Figure A-7.4 Figure A-7.5 Figure A-7.6 Figure A-7.7 Figure A-7.8 Figure A-7.9 Figure A-8.1 Figure A-8.2 Figure A-9.1 Figure A-10.1 Figure A-10.2 Figure A-11.1

Pressure-enthalpy diagram, Case study 11.1. . . . Automated HVAC control system architecture.. . Automated HVAC control system architecture.. . Pressure-enthalpy diagram, HFC-134a.. . . . . . Entropy. . . . . . . . . . . . . . . . . . . . . . . Mollier diagram, SI/Metric units. . . . . . . . . . Case study 7.1 Stage 1 sensible heat calculation. . Case study 7.1 Stage 2 latent heat calculation. . . Case study 7.1 Stage 3 sensible heat calculation. . Case study 7.1 Stage 4 latent heat calculation. . . Case study 7.1 Stage 5 sensible heat calculation. . Case study 7.1, Mass flow rate analysis. . . . . . Heat cycle in a rankine engine with superheat, h versus s. . . . . . . . . . . . . . . . . . . . . . Heat cycle in a rankine engine with superheat, h versus n.. . . . . . . . . . . . . . . . . . . . . Mollier diagram, self-assessment problem 1. . . . Psychrometric chart. . . . . . . . . . . . . . . . Psychrometric chart. . . . . . . . . . . . . . . . Pressure - enthalpy diagram, HFC-134a. . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

245 250 253 255 264 266 280 282 283 285 286 289

. .

292

. . . . .

293 297 304 307 309

. . . . .

List of Tables

Table 1.1 Table 1.2 Table 1.3 Table 1.4 Table 1.5 Table 1.6 Table 5.1 Table 5.2 Table 5.3 Table 5.4 Table 5.5 Table 5.6 Table 5.7 Table 6.1 Table 6.2 Table 6.3 Table 6.4 Table 6.5 Table 6.6

Units for pressure and associated conversion factors. Important temperatures and associated conversion factors. . . . . . . . . . . . . . . . . . . . . . . . . Rankine temperature conversion formulas. . . . . . . Kelvin temperature conversion factors. . . . . . . . . Approximate specific heat, cp, for selected liquids and solids, in kJ/kg °K, cal/gm °K, BTU/lbm °F, J/mol °K. . . . . . . . . . . . . . . . . . . . . . . . Densities of common materials. . . . . . . . . . . . Properties of saturated steam, by pressure, SI units. . . . . . . . . . . . . . . . . . . . . . . . . Superheated steam table excerpt, US/Imperial units. . . . . . . . . . . . . . . . . . . . Properties of saturated steam, by temperature, SI units. . . . . . . . . . . . . . . . . . . . . . . . . Properties of saturated steam, by pressure, US units. . Properties of saturated steam, by temperature, US units. . . . . . . . . . . . . . . . . . . . . . . . Properties of superheated steam, SI units. . . . . . . Properties of superheated steam, US units. n = specific volume in ft3/lbm; h = enthalpy in BTU/lbm; s = entropy in BTU/lbm-°R. . . . . . . . . Latent heats for water phase transformation, SI units. . . . . . . . . . . . . . . . . . . . . . . . . Properties of saturated steam by temperature, SI units. . . . . . . . . . . . . . . . . . . . . . . . . Properties of Saturated Steam By Pressure, US units. . . . . . . . . . . . . . . . . . . . . . . . Types of phase transformation. . . . . . . . . . . . . Critical properties of select substances. . . . . . . . . Triple point properties of select substances. . . . . .

xxiii

4 6 7 7 16 17 57 61 65 67 68 69 70 77 78 82 86 87 88

xxiv  List of Tables Table 7.1

Thermodynamic system definition, categorization and cross-referencing. . . . . . . . . . . . . . . . . . Table 7.1 Energy content and cost comparison of common fuels. . . . . . . . . . . . . . . . . . . . . . . . . . Table 7.2 Specific heat, cp, for selected liquids and solids. . . . Table 7.3 Latent heat for phase transformation of water. . . . . Table 7.4 Excerpt, superheated steam table, SI units. . . . . . . Table 7.4a Excerpt, saturated steam table, SI units . . . . . . . . Table 7.5 Energy content and cost comparison of common fuels. . . . . . . . . . . . . . . . . . . . . . . . . . Table 7.6 Specific heat, cp, for selected liquids and solids. . . . Table 7.7 Latent heat for phase transformation of water. . . . . Table 7.8 Superheated steam table excerpt, US/Imperial units. . . Table 7.9 Superheated steam table excerpt, US/Imperial units. . . Table 7.10 Saturated steam table excerpt, US/Imperial units. . . Table 7.11 Superheated steam table excerpt, US/Imperial units. . Table 7.12 Superheated steam table excerpt, US/Imperial units. . Table 8.1 Molecular weights of common gases associated with combustion reactions and byproducts.. . . . . . Table 11.1 Commonly used refrigerants and some of their important properties. . . . . . . . . . . . . . . . . . Table 11.2 Commonly used refrigerants and some of their important properties. . . . . . . . . . . . . . . . . . Table A-5.1 Properties of saturated steam, by pressure, SI units.. . . Table A-5.2 Superheated steam table excerpt, US/Imperial units. . . Table A-5.3 Properties of saturated steam, by temperature, SI units. . . . . . . . . . . . . . . . . . . . . . . . . Table A-5.4 Properties of saturated steam, by pressure, US units. . . Table A-6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Table A-6.2 Properties of saturated steam by pressure, US units. . . Table A-7.1 Superheated steam table excerpt, US/Imperial units. . . Table A-7.4a Excerpt, saturated steam table, SI units. . . . . . . . Table A-7.4 Excerpt, superheated steam table, SI units. . . . . . .

93 99 103 103 106 114 122 123 123 126 127 137 142 143 157 233 234 268 269 271 272 273 274 278 287 291

1 Introduction to Energy, Heat and Thermodynamics

Topics ■■

■■ ■■ ■■

Concepts, terms, principles, units, laws, and equations pertaining to energy. Heat and energy conversion. Specific heat. Energy transformation and associated case studies.

1.1 Introduction The term “thermodynamics” comes from two root words: “thermo,” which means heat, and “dynamic,” meaning energy in motion, or power. This also explains why the laws of thermodynamics are sometimes viewed as laws of “Heat Power,” where power represents the rate of flow of heat energy. Since heat is simply thermal energy, in this chapter, we will review energy basics and lay the foundation for in depth discussion on heat energy and set the tone for discussion on more complex topics in the thermodynamics realm.

1.2 Energy The capacity of an object, entity, or a system to perform work is called energy. Energy is a scalar physical quantity. In the International System of Units (SI), energy is measured in Newton-meters (Nm) or Joules, while in the US system of units, energy is measured in ft lbf, BTU’s, therms, or calories. In the field of electricity, energy is measured in watt-hours, (Wh), kilowatt-hours (kWh), Gigawatt-hours (GWh), Terawatt-hours (TWh), etc. Units for energy, such as ft-lbs and N-m, point to the equivalence of energy with torque (moment) and work, at the magnitude level. This point will be discussed later in this chapter. 1

2  Introduction to Energy, Heat and Thermodynamics Energy exists in many forms. Some of the more common forms of energy, and associated units, are as follows: 1.

Kinetic energy1; measured in ft lbf, BTUs, Joules, Nm (1 Nm = 1 Joule), etc. Where, BTU stands for British Thermal Units

2.

Potential energy1; measured in ft lbf, BTUs, Joules, Nm, etc.

3.

Thermal energy1, or heat (Q); commonly measured in Calories, BTUs, Joules, therms, etc.

4.

Internal energy1, (U); commonly measured in BTU’s, calories, or Joules.

5.

Electrical energy; measured in Watt-hours (Wh), killowatt-hours (kWh) and horsepower-hours (hp-h), etc.

6.

Gravitational energy; measured in ft lbf, Joules, Nm, etc.

7.

Sound energy; measured in Joules.

8.

Light energy; measured in Joules.

9.

Elastic energy; measured in ft lbf, BTUs, Joules, Nm, etc.

10. Electromagnetic energy; measured in Joules. 11. Pressure energy1; measured in ft lbf, BTUs, Joules, N m, etc. Note: These forms of energy are discussed in greater detail later in this chapter.

1

1.3 Root Concepts and Terms that Contribute toward the Production or Transformation of Energy 1.3.1  Force and mass Force or weight, in US or imperial units is measured in lbf. While, mass is measured in lbm. Mass, in lbm, is a scaler entity and can be converted to weight, in lbf, which is a vectoral entity, through multiplication by factor g/gc, as follows: Given: An object of mass, m = 1 lbm. Weight = Force = m(g/gc) = (1 lbm)(g/gc) = (1 lbm)(32.2 ft/s2)/(32.2 lbm ft/lbf s2) = 1 lbf

1.3 Root Concepts and Terms 3

Force or weight, in SI, or metric units, is measured in Newton’s, or “N”. Mass, in SI units, is measured in kg. Mass, available in kg, can be converted to weight or force, in Newton’s, through multiplication by, simply, the gravitational acceleration “g,” which is equal to 9.81 m/s2. Conversion of 1 kg of mass to its corresponding weight would be as follows: Given: An object of mass, m = 1 kg. Weight = Force = mg = (1 kg)(9.81 m/s2) = 9.81 kgm/s2 Note that since weight represents force, it is a vector and can be fully defined through statement of, both, magnitude, and direction. In this text, in most instances, vectoral entities are denoted through the use of bold fonts. Since, 1 N = 1 kgm/s2 Weight = 9.81 kgm/s2 = 9.81 N 1.3.2  Density and weight density Density is defined as mass per unit volume. The symbol for density is: ρ (Rho) Density of water, in the SI (Metric) realm, under STP conditions, is 1000 kg/m3. This density can be converted to specific weight, γ (gamma), or weight density, as follows: γ = ρ(g)  = (1000 kg/m3)(9.81 m/s2) = 9,810 N/m3

(1.1)

Density ρ of water, in the US (Imperial) realm, under STP conditions, is 62.4 lbm/ft3. The density can be converted to specific weight, γ, or weight density, as follows: γ = ρ(g) = (62.4 lbm/ft3)(g/gc) = (62.4 lbm/ft3)(32.2 ft/s2)/(32.2 lbm ft/lbf s2) = 62.4 lbf/ ft3 1.3.3  Specific volume Specific Volume is the inverse of density. The symbol for specific volume is “υ,” (upsilon). The formula for specific volume, υ, is: υ = 1/ρ(1.2)

4  Introduction to Energy, Heat and Thermodynamics Table 1.1  Units for pressure and associated conversion factors.

    1 Pa 1 bar

Pascal (Pa) 1 N/m2 100,000

bar (bar) 10−5 Approx. 106 dyn/cm2 1 atm 98,066.50 0.980665 1 atm 101,325 1 torr 133.322 1 psi

Units for Pressure atmos. (atm) 1.0197 × 10−5 1.0197

torr (Torr) 7.5006 × 10−3 750.06

Approx.1 kgf/cm2 735.56

1.01325 1.0332 1.3332 × 10−3 1.3595 × 10−3

6.894×103 68.948 × 10−3 70.307 × 10−3

psi (psi) 145.04 × 10−6 14.50377 14.223

760 14.696 Approx. 1 Torr; 19.337 × 10−3 or 1 mmHg 51.715 1 lbf/in2

The units for specific volume are as follows:

•• ••

US (Imperial) Units: υ is measured in cu ft/lbm or ft3/lbm. SI (Metric) Units: υ is measured in m3/kg.

1.3.4 Pressure Pressure is defined as force applied per unit area. The symbol used for pressure is p. The formula for pressure is: p = F/A

(1.3)

Where, F = Applied force A = Area over which the force is applied There are a number of units used, customarily, for measuring pressure. Common units for pressure, in the SI, or Metric system are N/m2, Pa (Pascals), kPa, or MPa. One Pascal is equal to 1 N/m2. Common unit for pressure, in the US, or Imperial, system is psi, or lbf/in2, psia (psi-absolute) or psig (psigage). Other units utilized for pressure, and corresponding conversions are shown in Table 1.1 below. 1.3.5 Temperature Temperature can be defined as a measure of the average kinetic energy of the particles in a substance, where such energy is directly proportional to the degree of hotness or coldness of the substance.

1.3 Root Concepts and Terms 5

While temperature is one of the principal parameters of thermodynamics, it must be clear that temperature is not a direct measurement of heat, Q. Temperature is, however, is a parameter that is instrumental in determining the direction of flow of heat, Q. In that, heat travels from bodies at higher temperature to bodies at lower temperature. This role of temperature comports with the laws of thermodynamics. From physics perspective, temperature is an indicator of the level of kinetic energy possessed by atoms and molecules in substances. In solids, at higher temperature, the atoms oscillate or vibrate at higher frequency and possess more energy. In atomic gases, the atoms, at higher temperatures, tend to exhibit faster translational movement. In molecular gases, the molecules, at higher temperatures, tend to exhibit higher rates of vibrational and rotational movement. Even though, for a system in thermal equilibrium at a constant volume, temperature is thermodynamically defined in terms of its energy (E) and entropy (S), as shown in Eq. 1.4 below, unlike pressure, temperature is not commonly recognized as a derivative entity and, therefore, the units for temperature are not derived from the units of other independent entities. T=

∂E (1.4) ∂S

The universal symbol for temperature is T. The unit for temperature, in the SI, or metric, realm is °C. In the Celsius temperature scale system, 0°C represents the freezing point of water. The unit for temperature, in the US, or imperial, realm is °F. On the Fahrenheit temperature scale system, 32°F represents the freezing point of water. The formulas used for conversion of temperature from metric to US realm, and vice and versa, are as follows:

(°C × ) + 32 = °F(1.5) 9

5

(°F - 32) x

5

9

= °C(1.6)

1.3.6  Absolute temperature Unlike the Celsius temperature scale system, where 0°C represents the freezing point of water, the absolute temperature scale defines temperature independent of the properties of any specific substance. According to the laws of thermodynamics, absolute zero cannot be reached, practically, because this would require a thermodynamic system to be fully removed from the rest of the universe. Absolute zero is the theoretical temperature at which

6  Introduction to Energy, Heat and Thermodynamics Table 1.2  Important temperatures and associated conversion factors.

  Absolute Zero

Important Temperatures and Conversion Table Kelvin Celsius Fahrenheit 0°K −273.15°C −459.67°F

Freezing Point of 273.15°K Water Triple Point of Water 273.16°K Boiling Point of Water 373.13°K

Rankine 0°R

0°C

32°F

491.67°R

0.01°C 99.98°C

32.0°F 211.97°F

491.69°R 671.64°R

entropy would reach its minimum value. Absolute zero is defined as 0°K on the Kelvin scale and as −273.15°C on the Celsius scale. This equates to −459.67°F on the Fahrenheit scale. . It is postulated that a system at absolute zero would possess finite quantum, mechanical, zero-point energy. In other words, while molecular motion would not cease entirely at absolute zero, the system would lack enough energy to initiate or sustain transference of energy to other systems. It would, therefore, be more accurate to state that molecular kinetic energy is minimal at absolute zero. According to the Second Law of Thermodynamics (discussed later in this text), at temperatures approaching the absolute zero, the change in entropy approaches zero. This comports with the stipulation that as temperatures of systems or bodies approach absolute zero, the transference of heat energy diminishes. Equation 1.7 is mathematical statement of the Second Law of Thermodynamics. lim ΔS = 0 (1.7) T →0

Scientists, under laboratory conditions, have achieved temperatures approaching absolute zero. As temperature approaches absolute zero, matter exhibits quantum effects such as superconductivity and superfluidity. A substance in a state of superconductivity has electrical resistance approaching zero. In superfluidity state, viscosity of a fluid approaches zero. Table 1.2 shows factors for conversion of temperatures between Kelvin, Celsius, Fahrenheit and Rankine scales. This table also shows absolute temperature, freezing point of water, triple point for water and the boiling point of water. Tables 1.3 and 1.4 list formulas for conversion of temperatures between Kelvin, Celsius, Fahrenheit and Rankine scales.

1.3 Root Concepts and Terms 7 Table 1.3  Rankine temperature conversion formulas.

  Celsius Fahrenheit Kelvin

Rankine Temperature Conversion Formulas From Rankine To Rankine [°C] = ([°R] − 492) × 5⁄9 [°R] = ([°C] + 273) × 9⁄5 [°F] = [°R] − 460 [°R] = [°F] + 460 [K] = [°R] × 5⁄9 [°R] = [K] × 9⁄5 Table 1.4  Kelvin temperature conversion factors.

  Celsius Fahrenheit Rankine

Kelvin Temperature Conversion Formulas From Kelvin To Kelvin [°C] = [K] − 273 [K] = [°C] + 273 [°F] = [K] × 9⁄5 − 460 [K] = ([°F] + 460) × 5⁄9 [°R] = [K] × 9⁄5 [K] = [°R] × 5⁄9

In the metric or SI system, the absolute temperature is measured in°K. The relationship between°C and°K is as follows: T°K = T°C + 273° and, ΔT°K = ΔT°C In the US system, the absolute temperature is measured in°R. The relationship between°F and°R is as follows: T°R = T°F + 460° and, ΔT°R = ΔT°F The absolute temperature system should be used for all thermodynamics calculations, unless otherwise required. 1.3.7  Law of conservation of energy The law of conservation of energy states that energy can be converted from one form to another but cannot be created or destroyed. This can be expressed, mathematically, as: ∑ E = ∑ Energy = Constant

8  Introduction to Energy, Heat and Thermodynamics

1.4 Forms of Energy in Mechanical and Thermodynamic Systems 1.4.1  Potential energy Potential energy is defined as energy possessed by an object by virtue of its height or elevation. Potential energy can be defined, mathematically, as follows: Epotential = mgh,

{SI Units}

(1.8)

Epotential = m(g/gc)h,

{US Units}

(1.8a)

When the change in potential energy is achieved through performance of work, W: W = D Epotential 

(1.9)

1.4.2  Kinetic energy Kinetic energy is defined as energy possessed by an object by virtue of its motion. Kinetic energy can be defined, mathematically, as follows: Ekinetic = ½mv2

{SI Units}

(1.10)

Ekinetic = ½ (m/gc) v2

{US Units}

(1.10a)

Where, m = mass of the object in motion v = velocity of the object in motion gc = 32 lbm ft/lbf s2 When the change in kinetic energy is achieved through performance of work, W: W = D Ekinetic 

(1.11)

1.4.3  Energy stored in a spring2 Potential energy can be stored in a spring - or in any elastic object - by compression or extension of the spring. Potential energy stored in a spring can be expressed, mathematically, as follows:

1.4  Forms of Energy in Mechanical and Thermodynamic Systems  9

Espring = ½kx2  And,

(1.12)

Wspring = D Espring (1.13)

Where, k = The spring constant x = The contraction or expansion of the spring Note: In steel beam systems, beams act as springs, when loaded, to a certain degree. The deflection of a beam would represent the “x,” in Eq. 1.11.

2

1.4.4  Pressure energy Energy stored in a system in form of pressure is referred to as pressure energy. For instance, energy stored in a compressed air tank is pressure energy. Pressure energy can be expressed, mathematically, as follows: Epressure = Eflow = m p υ (1.14) Where, m = mass of the pressurized system; this would be compressed air in a compressed air system p = pressure in the system υ = is the specific volume 1.4.5  Heat and internal energy of a system According to the law of conservation of energy, if heat “Q” is added or removed from a system, in the absence of net work performed by or on the system, change in the internal energy “U “of a system would be: Uf – Ui = DQ(1.15) Where Q is positive when heat flows into a thermodynamic system and it is negative when heat exits a system. Specific internal energy “u” is defined as internal energy per unit mass. The units for internal energy are BTU/lbm, in the US System, and are kJ/kg, in the Metric or SI System.

10  Introduction to Energy, Heat and Thermodynamics 1.4.6  Unit conversions3 associated with heat energy Some of the common heat energy units and unit conversion formulas are listed below:

••

Conversion of heat energy measured in MMBTU’s to BTU’s:

○ 1 MMBTU × (1000,000 BTU/MMBTU) = 10

6

••

MMBTU’s

Conversion of the rate of heat energy transfer, measured in BTU’s/hr to tons:

○ 12,000 BTU/hr = 1 ton of refrigeration capacity = 3.52 kW

••

Conversion of heat energy measured in Deca Therms to BTU’s:

••

Conversion of heat energy measured in BTU’s to kWh and KWh to BTU’s:

○ 1 dT × (1,000,000 BTU/dT) = 1,000,000 BTU’s or 1MMBTU ○ 1 BTU × (2.928 × 10 kWh/BTU) = 0.0002928 kWh ○ 1 kWh × (3413 BTU/ kWh) = 3,413 BTU’s –4

Note: These heat energy conversion formulas will be used in various analysis and example problems through this text.

3

1.4.7  Molar internal energy Molar internal energy “U” is defined as internal energy per mole. The units for internal energy U are BTU/lbmole, in the US system, and are kJ/kmole, in the Metric or SI System.

1.5  Case Study 1.1: Energy and Energy Unit Conversion As an energy engineer, you are to analyze substitution of coal, as heating fuel, in lieu of nuclear energy derived from complete conversion, of 2.5 grams of a certain mass. The nuclear reaction is similar to the Uranium fission reaction shown in Figure 1.1 below. If the heating value of coal is 13,000 BTU/lbm, how many U.S. tons of coal must be burned in order derive the same amount of energy? Solution Given or known: c = Speed of light = 3 × 10 8 m/s m = Mass of material to be converted to Energy: 2.5 g, or 0.0025kg

1.5 Case Study 1.1 11

Figure 1.1  Uranium fission reaction.

Energy content of coal: 1 lbm of coal contains 13,000 BTU’s of energy Mass conversion factor; lbm to US tons, and vice and versa: 2000 lbm/ ton Energy unit conversion between Joules and BTU’s: 1055 Joules/BTU According to Einstein’s Equation: E = mc2 

(1.16)

By applying Eq. 1.16, Energy derived from 2 grams of given mass would be: E = mc2 = (0.0025 kg) × (3 × 10 8 m/s)2 = 2.25 × 1014 Joules This energy can be converted into BTU’s as follows: E = (2.25 × 1014 Joules) / ( 1055 Joules/BTU) = 2.13x 1011 BTU’s. Since 1 lbm of coal contains 13,000 BTU’s of heat, the number of lbs of coal required to obtain 2.13x 1011 BTU’s of heat energy would be: E = (2.13 × 1011 BTU’s ) / (13,000 BTU’s/lbm) = 1.64 × 1007 lbm

12  Introduction to Energy, Heat and Thermodynamics Since there are 2000 lbm per ton: E = (1.64 × 1007 lbm)/(2000 lbm/ton) = 8,203 US tons of coal. Conclusion: Energy derived from fission of 2.5 grams of fissile material is equivalent to the energy derived from 8,203 tons of coal. 1.5.1 Work As we will see, through the exploration of various topics in this text, work can be viewed as a vehicle for converting energy contained in various types of fuels to mechanical or electrical energy. In this section, we will elaborate on aspects of work that will be applied in the discussion and analysis of thermodynamic systems. 1.5.2  Work in a mechanical system In a mechanical system, work performed by an external force is referred to as external work. While, work performed by an internal force is referred to as internal work. Units such as BTU’s or kilocalories are not, customarily, used to measure mechanical work. In a mechanical system, work is positive when it is the result of force acting the direction of motion. Work is negative when it the result of a force opposing motion. Work attributed to friction is an example of negative work. Where, friction “Ff” is defined, mathematically, as: Ff = μf N 

(1.17)

Where, μf = Coefficient of friction N = Normal force applied by the surface against the object Frictional force, Ff, can either be static frictional force or dynamic frictional force. As stipulated by Eq. 1.17, Ff is directly proportional to the normal force N. The coefficient of friction for static friction is, typically, higher as compared to the coefficient of friction for dynamic friction. This is ostensible from the fact that greater force is required to set an object in motion as opposed to the force required to maintain the object in motion.

1.5 Case Study 1.1 13

Figure 1.2  Illustration of mechanical work, in a system with friction.

1.5.3  Mathematical equations for work Work can be performed, defined and computed in several ways. Some of the diferences stem from the realm or frame of reference that work occurs in. Listed below are some of the scenarios in which work can occur, and pertinent formulas: In a rotational system, with variable torque: Wvariable torque = ∫  T . dθ(1.18) In a rotational system, with constant torque: Wconstant torque = τ q 

(1.19)

Where, T = τ = Torque q = Angular distance traversed in the same direction as the torque, τ. In a linear system, with variable force: Wvariable force = ∫  F . ds(1.20) In a linear system, with constant force, where, force and distance are colinear: Wconstant force = F s. 

(1.21)

Note in Figure 1.2, the applied force F is not colinear with the line of motion, only its component F.Cosθ is.

14  Introduction to Energy, Heat and Thermodynamics General equation for work performed by a constant frictional force where, force and distance are colinear: W friction = Ff s

(1.22)

When work is performed by a force, F, that is applied at an angle, θ, with respect to the direction of motion – as shown in Figure 1.2 - it can be defined, mathematically, as follows: W = Work = (F Cosθ - Ff ) s 

(1.23)

W = (F Cosθ) s – Ff s 

(1.24)

Or,

Where, s = Distance over which the force is applied In Equations 1.23 and 1.24, the mathematical term “(FCosθ) s” constitutes positive work performed by the force F in the direction of motion, and (Ff s) constitutes negative work, performed against the direction of motion. Note that component “F.Cosθ,” in Equations 1.23 and 1.24, represents the horizontal component of force contributed by the diagonally applied force F. Work performed by gravitational force is defined, mathematically, as: Wg (SI) = mg(hf - hi), in the Metric Unit Systems 

(1.25)

Wg(US) = m(g/gc) (hf - hi), in the US Unit Systems 

(1.26)

Where, hf = The final elevation of the object hi = The initial elevation of the object Work performed in the case of a linear spring expansion or contraction is represented, mathematically, as: Wspring = ½k (xf – xi)2  Where, k = The spring constant xi = The initial length of the spring xf = The final length of the spring

(1.27)

1.5 Case Study 1.1 15

1.5.4  Work performed in a thermodynamic system In the thermodynamics domain, work constitutes the phenomenon of changing the energy level of an object or a system. The term “system,” in thermodynamics, is often used interchangeably with the term medium. For instance, in the case of an open thermodynamic system - such as steam powered turbine - steam is considered as a system performing work on the surroundings, i.e., the turbine. In a thermodynamic system, work is positive when an object or system performs work on the surroundings. Example: If the vanes of an air compressor are considered to constitute the system, then the work performed on air, by the vanes in an air compressor, would be positive. Work is negative when the surroundings perform work on the object. Inflating of a raft or an inner tube constitutes negative work as the air (environment or surrounding) performs work on the walls of the raft or tube (the system) during the inflation process. 1.5.5  Specific heat Specific heat is defined as the amount of the heat, Q, required to change the temperature of mass “m” of a substance by ΔT. The symbol for specific heat is “c.” The mathematical formula for specific heat of solids and liquids is: c = Q /(mΔT) 

(1.28)

Q = mcΔT 

(1.29)

Or,

Where, m = Mass of the substance; measured in kg, in the SI system, and in lbm in the US system Q = The heat added or removed; measured in Joules or kJ in the SI System, or in BTU’s in the US system ΔT = The change in temperature, measured in°K in the SI Systems, or in°R in the US System The units for c are kJ/(kg.°K), kJ/(kg.°C), BTU/(lbm.°F) or BTU/(lbm.°R).

16  Introduction to Energy, Heat and Thermodynamics The thermodynamic equation involving specific heats of gases are as follows: Q = mcv.ΔT, when volume is held constant.

(1.30)

Q = m cp. ΔT, when pressure is held constant.

(1.31)

Approximate specific heat, cp, for selected liquids and solids are listed in Table 1.5, below. The next case study, Case Study 1.2, is designed to expand our exploration of energy related analysis methods and computational techniques. Some of the energy, work and heat considerations involved in this case study lay a foundation for more complex energy work and thermodynamics topics that lie ahead in this text. This case study also provides us an opportunity to experience the translation between the SI (Metric) unit system and the US (Imperial) unit system. As we compare the solutions for this case study in the US and SI unit systems, we see that choosing one unit system versus another, in some cases does involve the use of different formulas. This difference in formulas for different unit systems is evidenced in the potential and kinetic energy components of the energy conservation equations. Table 1.5  Approximate specific heat, cp, for selected liquids and solids, in kJ/kg °K, cal/g °K, BTU/lbm °F, J/mol °K.

Substance Aluminum Bismuth Copper Brass Gold Lead Iron Silver Tungsten Zinc Mercury Ethyl Alcohol Water Ice at -10 °C Granite Glass

cp in kJ/kg °K 0.9 0.123 0.386 0.38 0.126 0.128 0.460 0.233 0.134 0.387 0.14 2.4 4.186 2.05 0.79 0.84

cp in kcal/kg °K or BTU/lbm °F 0.215 0.0294 0.0923 0.092 0.0301 0.0305 0.11 0.0558 0.0321 0.0925 0.033 0.58 1 0.49 0.19 0.2

Molar Cp J/mol °K 24.3 25.7 24.5 N/A 25.6 26.4 N/A 24.9 24.8 25.2 28.3 111 75.2 36.9 N/A N/A

1.6 Case Study 1.2

17

Table 1.6  Densities of common materials.

Metal Water Aluminum Zinc Iron Copper Silver Lead Mercury Gold

Densities of Common Materials lb/in3 lb/ft3  g/cm3 1.00  0.036  62 2.7 0.098 169 7.13 0.258 445 7.87 0.284 491 8.96 0.324 559 10.49 0.379 655 11.36 0.41 709 13.55 0.49 846 19.32 0.698 1206

lb/gal  8.35  22.53 59.5 65.68 74.78 87.54 94.8 113.08 161.23

Furthermore, this case study helps us understand the vital and integrated role that work, kinetic energy and potential energy play in application of law of conservation of energy in thermodynamic system analysis. In this case study we start off with energy and work considerations and conclude our analysis with the quantitative assessment of thermodynamic impact on steel, air and water, key substances involved in the overall process.

1.6 Case Study 1.2: Energy Conservation, Energy Conversion and Thermodynamics At a foundry, a solid rectangular block of carbon steel, density 7,850 kg/ m3 (491 lbm/ft3), from Table 1.6, is released to a downward inclined ramp as shown in Figure 1.3. The volume of the block is 1.0 m3 (35.32 ft3) and its release velocity, at the top of the ramp is 1.5 m/s (4.92 ft/s). The force of friction between the block, the inclined surface and the flat conveyor bed is assumed to be a constant amount of 400 N (89.924 lbf) for simplicity (of course, in a real scenario the force of friction will be different between the inclined paths and the flat ones). The block is stopped on the flat section through compression of a shock absorbing spring system before it settles on a roller conveyor operating at a linear speed of 2 m/s (6.562 ft/s). Assume that the frictional force stays constant through the entire path of the block. a) Determine the velocity of the steel block when it enters the horizontal segment of the travel, i.e., point “w.” b) Employing the law of energy conservation and principles of energy conversion, calculate the value of the spring constant for the shock absorbing spring system.

18  Introduction to Energy, Heat and Thermodynamics

Figure 1.3  Case Study 1.2.

c) What would the steady state speed of the block be after it settles onto the roller conveyor? d) If the spring type shock absorbing system is replaced by a compressed air cylinder of 1.0 m3 (35.32 ft3) uncompressed volume, at room temperature of 20°C (68°F) and standard atmospheric pressure, what would be the rise in temperature of the cylinder air immediately after the steel block’s impact? The final, compressed, volume is 0.75 m3 (26.49 ft3), and the pressure gage on the cylinder reads 2bar (29 psia). f) If the conveyor, at the bottom of the incline, is a belt driven roller conveyor and the rate of flow of blocks onto the conveyor is one per 10 ­seconds, determine the horsepower rating of the conveyor motor. Assume the conveyor belt to be directly driven off the conveyor motor shaft and that there is no slip between the belt and the rollers. Assume the motor efficiency to be 90%. g) The conveyor transports the blocks to a cooling/quenching tank. The temperature of the blocks, when they are dropped into the quenching tank, is 100°C (212°F). The initial temperature of the water in the quench tank is 20°C (68°F) and volume of water is 6.038 m3 (213.23 ft3). The final, equilibrium, temperature of the water and the block is 30°C (86°F). Determine the amount of heat extracted by the quench water per block.

1.6 Case Study 1.2 19

Solution, Case Study 1.2 – SI (Metric) version Given, known or derived values: = Vsteel block = 1.0 m3 Volume of the steel block Density of the steel block = rsteel = 7850 kg/m3 Mass of the steel block = msteel block = rsteel Vsteel block = (7850 kg/m3 ) × (1.0 m3 ) \ msteel block = 7850 kg Velocity of block at point “z” = Vz = 1.5 m/s Elevation at point “z” = hz = 2.0 m Spring constant for the shock absorbing spring = k Frictional force, Ff = N. µf = 400 N Density of air at STP: 1.29 kg/ m3 At STP:   Temperature = 0°C,   Pressure = 1 bar, 101.33kPa, or 101,330 Pa a) Determine the velocity, vw, of the steel block when it enters the horizontal segment of the travel, i.e. point w: Solution Strategy: The velocity variable vw is embedded in the kinetic energy of the block, 1/2 m. vw2. So, if we can determine the amount of kinetic energy possessed by the block at point w, we can derive the required velocity vw. To find the kinetic energy at point w, we can apply the law of conservation of energy at points z and w as shown below: Let total energy at point z = Ez-total Then, Ez-total = Ez-kinetic + Ez-potential Ez-total = 1/2 m. vz2 + m g hz Ez-total = 1/2 (7850 kg) (1.5 m/s)2 + (7850 kg) (9.81 m/s2 ) (2 m)   ∴ Ez-total = 162,848 J The energy lost in the work performed against friction, during the block’s travel from z to w, is accounted for as follows: W f- wz = Work performed against friction = (Dist. w–z ) (Ff ) = (50 m) (400 N) = 20,000 J

20  Introduction to Energy, Heat and Thermodynamics Therefore, the energy left in the block when it arrives at point w, at the bottom of the ramp, would be as follows: Ew-total = Ez-total - W f- wz = 162,848 J - 20,000 J = 142,848 J Since the block is at “ground” elevation when it arrives at point w, the potential energy at point w would be “zero.” \ Ew-total = 1/2 mvw2 Or, Vw = 2

E w - total m

 148, 848 J  = 2  7, 850 kg  = 6.03 m/s b) Employing the law of energy conservation and principles of energy conversion, calculate the value of the spring constant for the shock absorbing spring system. Solution Strategy: The unknown constant k is embedded in the formula for the potential energy stored in the spring after it has been fully compressed, upon stopping of the block. This potential energy is equal to the work performed on the spring, i.e., W spring = 1/2kx2. So, if we can determine the amount of work performed on the spring, during the compression of the spring, we can derive the required value of k. To derive the value of Wspring, we will apply the law of conservationof energy to the travel of the block from point z to point x. Based on the dimensions in Figure 1.3: Distance x-z = (0.55m + 50 m + 40m) = 90.55 m W f - xz = Work performed against friction over Dist. x-z = (Dist. x-z )Ff = (90.55 m)(400 N) = 36,220 J

1.6 Case Study 1.2 21

Therefore, the total energy at point z would be: Ez-total = W spring + W f - xz Or, W spring = Ez-total - W f - xz = 162,848 J – 36,220 J = 126, 628 J Energy stored in the spring is quantified as: W spring = 1/2 kx2 Since W spring has beed determined to be equal to 126,628 J, 126,628 J = ½k(0.55) 2 Therefore, k = 2 Wspring / x2 = 2 Wspring / x2 = 2 (126,628 J) / (0.55) 2 = 837,212 N/m c) What would the steady state speed of the block be after it settles onto the roller conveyor? Solution/answer: After the block settles into a steady state condition on the conveyor, it assumes the speed of the conveyor, i.e., 2 m/s. d) Rise in the temperature of the compressed air in the shock absorbing cylinder: Solution: The rise in the cylinder’s air temperature can be determined after calculating the final temperature of the air through the application of the ideal gas law. Ideal gas laws can be applied in this case because air, for most practical purposes, is assumed to act as an ideal gas. Also, note that air – which is a composition, mainly of nitrogen and oxygen – is in a superheated state under STP conditions. This means that the gases that air is composed of, under STP conditions, are well beyond their saturated vapor temperature and pressure conditions. For a gas to be treated as an ideal gas, it must exist in superheated state. This will be elucidated in greater detail in later chapters.

22  Introduction to Energy, Heat and Thermodynamics According to ideal gas law: (P1V1)/ T1 = (P2V2)/T2 

(1.32)

Or, through rearrangement of Eq. 1.32: T2 = (P2V2T1 )/(P1V1 ) 

(1.33)

Given or known: P1 = 1 Bar = 101.33 kPa V1 = 1.0 m3 T1 = 20°C => 273 + 20°C = 293°K P2 = 2 bar or 202.66 kPa V2 = 0.75 m3 Then, by applying Eq. 1.33:

(202.66 kPa ) (0.75 m ) (293°K ) (101.33 kPa ) (1.0 m ) 3

T2 =

3

= 439.5 °K i.e. 166.5°C; Therefore, the rise in the cylinder air temperature would be: = 166.5°C - 20°C = 146.5°C Note: In Chapter 8, this part, (e), will be extended, as an illustration of isothermal process, to determine the amount of heat that must be removed to maintain the air temperature at 20°C. e) If the conveyor, at the bottom of the incline, is a belt driven roller conveyor and the rate of flow of blocks onto the conveyor is one per 10 ­seconds, determine the horsepower rating of the conveyor motor. Assume the conveyor belt to be directly driven off the conveyor motor shaft and that there is no slip between the belt and the rollers. Assume the motor efficiency to be 90%. Solution: Apply the power, velocity and force formula to determine the power requirment, as follows: P = Fv 

(1.34)

1.6 Case Study 1.2 23

Where, P = Power Required to move the steel blocks F = Force required to move the block or the force required to move the conveyor belt with the block on the rollers v = Velocity of the belt; i.e. 2.0 m/s, as given. While the velocity v is given, the force F is unknown and must be derived. Force can be defined in terms of mass flow rate ṁ and the change in velocity Δv, as stated in Eq. 1.35 below: F = ṁ Δv

(1.35)

Based on the derived mass of the block as 7850 kg and the fact that one block is moved every 10 seconds: ṁ = mass flow rate = 7850 kg / 10 s = 785 kg/s And, based on the given conveyor speed of 2.0 m/s: = Change in the velocity of the block Δv = vf - vf = 2.0 m/s – 0 \ Δv = 2.0 m/s {Note: This change in the velocity is in the direction of the roller conveyor} Therefore, applying Eq. 1.35, the force required to move the block would be: F = ṁ Δv = (785 kg/s ) (2.0m/s – 0) = 1570 N {Note: (kg/s m/s) => kg m/s 2 m a} Then, by applying Eq. 1.34: P =Fv = (1570 N)(2 m/s) = 3140 W Since there are 746 watts per hp, the computed power of 3140 W, in hp, would be: P = (3140 W) /(746 W/hp) = 4.21 hp

24  Introduction to Energy, Heat and Thermodynamics Therefore, choose a standard 5 hp motor. Note: The efficiency of the motor is not needed in the motor size determination. Motor rating or size is specified on the basis of the brake horsepower required by the load; which in this case, is 4.21hp, rounded up to 5hp. f) The conveyor transports the blocks to a cooling/quenching tank. The temperature of the blocks, when they are dropped into the quenching tank, is 100°C. The initial temperature of the water in the quench tank is 20°C and volume of water is 6.038 m3. The final, equilibrium, temperature of the water and the block is 30°C. How much heat is extracted by the quench water per block? Solution Given or known: c cast iron = 0.460 kJ/kg.°K msteel block = 7850 kg, as determined earlier = 100°C = 273 + 100°C = 373°K Tblock - i Tblock - f = 30°C = 273 + 30°C = 303°K ∴ ΔTblock = 303°K - 373°K= - 70°K

{From Table 1.5}

According to Eq. 1.29: Q = mcΔT Therefore, Q lost by the block = (m block)(c )(ΔTblock) Q lost by the block = (7850 kg)(0.460 kJ/kg°K )(-70°K) = - 252,770 kJ Since, Q absorbed by water = - Q lost by the block, Q absorbed by water = - (- 252,770 kJ ) = + 252,770 kJ Solution, Case Study 1.2 – US (Imperial) Unit Version Given, known or derived values: Volume of the steel block = Vsteel block = 35.32 ft3 Density of the steel block = rsteel = 491 lbm/ft3 Mass of the steel block = msteel block = rsteel Vsteel block = (491 lbm/ft3) × (35.32 ft3)

1.6 Case Study 1.2 25

\ msteel block = 17,342 lbm Velocity of block at point “z” = Vz = 4.92 ft/s Elevation at point “z” = hz = 6.56 ft Spring constant for the shock absorbing spring = k Frictional force, Ff = N. µf = 89.92 lbf Density of air at STP: 0.0805 lbm/ft3, at STP: Temperature = 32°F, Pressure = 1 atm or 14.7 psia a) Determine the vvelocity, vw, of the steel block when it enters the horizontal segment of the travel, i.e. point w: Solution Strategy: The velocity variable vw is embedded in the kinetic energy of the block: 1/2m/gcvw2. So, if we can determine the amount of kinetic energy possessed by the block at point w, we can derive the required velocity vw. To find the kinetic energy at point w, we can apply the law of conservation of energy at points z and w as shown below: Let total energy at point z = Ez-total Then, Ez-total Ez-total Ez-total ∴ Ez-total

= Ez-kinetic + Ez-potential = 1/2(m /gc)vz2 + m(g/gc)hz  = 1/2 {(17,342 lbm/(32 lbm ft/lbf s2 )} (4.92 ft/s)2 + (17,342 lbm )(32 ft/s2/32 lbm ft/lbf s2 )(6.56 ft) = 120,282 ft lbf

(1.36)

The energy lost in the work performed against friction, during the block’s travel from z to w, is accounted for as follows: W f- wz = Work performed against friction = (Dist. w-z )(Ff ) = (164 ft)(89.92 lbf) = 14,747 ft lbf Therefore, the energy left in the block when it arrives at point w, at the bottom of the ramp, would be as follows: Ew-total

= Ez-total - W f- wz = 120,282 ft lbf - 14,747 ft lbf = 105,535 ft lbf

26  Introduction to Energy, Heat and Thermodynamics Since the block is at “ground” elevation when it arrives at point w, the potential energy at point w would be “zero.” \ Ew-total = 1/2 (m/ gc) vw2 (1.37) Or,

vw = 2 gc

E w-total m

v w = 2 (32lbm ft / lbf s 2 )

105, 535 ft lbf = 19.74 ft/s 17, 342 lbm

b) Employing the law of energy conservation and principles of energy conversion, calculate the value of the spring constant for the shock absorbing spring system. Solution Strategy: The unknown constant k is embedded in the formula for the potential energy stored in the spring after it has been fully compressed, upon stopping of the block. This potential energy is equal to the work performed on the spring, i.e., W spring = 1/2kx2. So, if we can determine the amount of work performed on the spring, during the compression of the spring, we can derive the required value of k. To derive the value of Wspring, we will apply the law of conservationof energy to the travel of the block from point z to point x. Based on the dimensions in Figure 1.3: Distance x-z = (1.8 ft + 131.23 ft + 164 ft) = 297 ft W f - xz

= Work performed against friction over Dist. x-z = (Dist. x - z)Ff = (297 ft)(89.92 lbf) = 26,706 ft lbf

Therefore, the total energy at point z would be: Ez-total = W spring + W f - xz

1.6 Case Study 1.2 27

Or,

W spring = Ez-total - W f - xz = 120,282 ft lbf – 26,706 ft lbf = 93,576 ft lbf

Energy stored in the spring is quantified as: Wspring = 1/2kx2 Since Wspring has beed determined to be equal to 93,576 ft lbf, 93,576 ft lbf = ½k(1.8 ft) 2 Therefore, k = 2 Wspring / x2 = 2 Wspring / x2 = 2 (93,576 ft lbf )/(1.8 ft) 2 = 57,763 lbf/ft Ancillary Exercise: This value of k = 57,763 lbf/ft, in US units, is within 0.7% of the value of k = 837,212 N/m, derived in SI units. The reader is encouraged to perform the unit conversions necessary to prove the practical equivalence between the k values calculated in US and SI units. c) What would the steady state speed of the block be after it settles onto the roller conveyor? Solution/answer: After the block settles into a steady state condition on the conveyor, it assumes the speed of the conveyor, i.e., 6.56 ft/s. d) Rise in the temperature of the compressed air in the shock absorbing cylinder: Solution: The rise in the cylinder’s air temperature can be determined after calculating the final temperature of the air through the application of the ideal gas law. Ideal gas laws can be applied in this case because air, for most practical purposes, is assumed to act as an ideal gas. According to ideal gas law: (P1 V1)/ T1 = (P2 V2)/ T2 

(1.32)

28  Introduction to Energy, Heat and Thermodynamics Or, through rearrangement of Eq. 1.32: T2 = (P2 V2T1 )/(P1V1) 

(1.33)

Given or known: P1 = 1 Atm = 14.7 psia V1 = 35.32 ft3 T1 = => 461 + 68°F = 529°R P2 = 2 Atm. = 29 psia V2 = 26.49 ft3 Then, by applying Eq. 1.33: T2 = {(29 psia)(26.49 ft3)( 529°R )}/{(14.7 psia) (35.32 ft3)} T2 = 782.7°R Or, T2 = 793.5°R - 461 = 332.5°F Therefore, the rise in the cylinder air temperature would be: = 332.5°F - 68°F = 264.5°F e) If the conveyor, at the bottom of the incline, is a belt driven roller conveyor and the flow of blocks onto the conveyor is one per 10 seconds, determine the horsepower rating of the conveyor motor. Assume the conveyor belt to be directly driven off the conveyor motor shaft and that there is no slip between the belt and the rollers. Assume the motor efficiency to be 90%. Solution: Apply the power, velocity and force formula to determine the power requirment, as follows: P = Fv

(1.34)

Where, P = Power Required to move the steel blocks F = Force required to move the block or the force required to move the conveyor belt with the block on the rollers v = Velocity of the belt; i.e. 6.562 ft/s, as given.

1.6 Case Study 1.2 29

While the velocity v is given, the force F is unknown and must be derived. Force can be defined in terms of mass flow rate ṁ and the change in velocity Δv, as stated in Eq. 1.38 below: F = (ṁ/gc) Δv(1.38) Based on the derived mass of the block as 7850 kg and the fact that 1 block is moved ecery 10 seconds: ṁ = mass flow rate = 17,342 lbm / 10 s = 1,734 lbm/s And, based on the given conveyor speed of 6.562 ft/s: Δv = Change in the velocity of the block = vf - vf = 6.562 ft/s – 0 \ Δv = 6.562 ft/s {Note: This change in the velocity is in the direction of the roller conveyor} Then, by applying Eq. 1.38, the force required to move the block would be: F = (ṁ/gc)Δv = {(1,734 lbm/s)/( 32 lbm ft/lbf s2)}(6.562 ft/s) = 356 lbf Then, by applying Eq. 1.34: P = Fv = (356 lbf)(6.562 ft/s) = 2336 ft lbf/s In hp, the computed power of 2336 ft lbf/s would be: P = (2336 ft lbf/s )/(550 ft lbf/s/hp) = 4.25 hp Therefore, choose a standard 5 hp motor. Note: The efficiency of the motor is not needed in the motor size determination. Motor is specified on the basis of the brake horsepower required by the load; which in this case, is 4.21hp.

30  Introduction to Energy, Heat and Thermodynamics f) The conveyor transports the blocks to a cooling/quenching tank. The temperature of the blocks, when they are dropped into the quenching tank, is 212°F. The initial temperature of the water in the quench tank is 68°F and volume of water is 213.23 ft3. The final, equilibrium, temperature of the water and the block is 86°F. Determine the amount of heat extracted by the quench water per block? Solution: Given or known: c cast iron = 0.11 BTU/lbm°F   or 0.11  BTU/lbm°R {From Table 1.5} msteel block = 17,342 lbm, as determined earlier. Tblock - i = 212°F => 461 + 212°F = 673°R Tblock - f = 86°F => 461 + 86°F = 547°R ∴ ΔTblock = 673°R - 547°R = 126°R And,

ΔTblock = 212°F - 86°F = 126°F

According to Eq. 1.29: Q = mcΔT Eq. 1.29 Therefore, Q lost by the block = (m block)(c )(ΔTblock) Q lost by the block = (17,342 lbm) (0.11BTU/lbm°R) (-126°R) = - 240,360 BTU Since, Q absorbed by water = - Q lost by the block, Q absorbed by water = - (-240,360 BTU ) = + 240,360 BTU

Chapter 1 Self-Assessment Problem and Question 1.

Determine the amount of heat extracted by the quench water, per block, in Case Study 1.2, using the temperature rise of the water when the steel block is dropped into the quenching tank. The temperature of the block is 100°C when it enters the quench water. The initial temperature of the water in the quench tank is 20°C and volume of water is 6.038 m3. The final, equilibrium, temperature of the water and the block is 30°C.

2 Thermodynamics and Power

Topics ■■ ■■ ■■

Concepts of power, power conversion, and efficiency Steam to wire power and energy transformations Steam to electrical power case study

2.1 Introduction This text focuses not only on the important concepts, theories, principles, and analyses techniques associated with thermodynamics but also demonstrates their practical applications through case studies that illustrate the flow of energy from thermal form to utilities such as electricity. Through some of the later chapters in this text we will learn how energy is harvested from fuels, fossil or non-fossil, transferred to a medium like water and packaged in form of the medium’s enthalpy – or enthalpy of the superheated steam. In subsequent chapters of this text, we will get an opportunity to study the concepts of enthalpy, entropy, and work in greater depth. We will also learn how the superheated steam transfers its heat energy to turbines and how the turbines transform the thermal energy (enthalpy) into electrical energy. However, to understand practical thermodynamic systems in a comprehensive fashion in the chapters ahead, in this chapter, we will examine what happens to the energy after it is transformed from enthalpy to work performed by the turbine. Since thermodynamic systems are constructed and installed for applications and purposes that extend beyond the boilers and turbines, knowledge and appreciation of flow of energy downstream of the turbines is essential for ensuring that investment in comprehensive power generating systems, as a whole, is productive and effective. This brief chapter prepares us to better understand the flow of energy beyond the turbines, through examination of energy as it transforms and flows from superheated steam enthalpy form to electrical power delivered onto to the electrical power grid. 31

32  Thermodynamics and Power Before we embark on the exploration of flow of energy, let’s review the concepts of power and efficiency.

2.2  Power and Efficiency 2.2.1 Power The concept of power was introduced briefly in the last chapter. We introduced the fact that power is rate of performance of work, or P = Work/ Elapsed Time and that is one of the ways power can be calculated is through the mathematical relationship Power = Force × Velocity. The counterpart of the last power formula, in the rotational motion realm, would be Power = Torque × Rotational Velocity. Other aspects of power, forms of power and formulas for power will be introduced and discussed, in depth, in subsequent chapters of this text. 2.2.2  Units for power US/Imperial Unit System: hp, ft-lbf/s, ft-lbf/min, BTU/s SI or Metric: Watts, kW, MW, GW, TW (10 12 W) 2.2.3  Common power conversion factors in the SI system 1 J/s = 1 Nm/s = 1 W 1 kJ/s = 1 kW 1000 kW = 1 MW 1.055 kJ/s = 1 BTU/s One hp = 746 Watts = 0.746 kW = 550 ft-lbf/s Since the units for power and energy are often confused, let’s also examine common units for energy so that the similarities and differences between the units for power and energy can be observed and noted. 2.2.4  Units for energy US/Imperial Unit System: ft-lbf, BTU 2.2.5  SI or Metric unit system Nm, Joules or J, Wh, kWh, MWh, GWh, TWh (10 12 Wh)

2.2  Power and Efficiency  33

2.2.6  Common energy conversion factors 1 J = 1 Nm 1 W × 1h = 1 Wh 1 kW × 1h = 1 kWh 1000 kW × 1h = 1 MWh 1 BTU = 1055 J = 1.055 kJ 1 BTU = 778 ft lbf 1 hp × 1hour = 1 hp-hour 2.2.7 Efficiency Efficiency is defined, generally, as the ratio of output to input. The output and input could be in form of power, energy or work. Efficiency assumes a more specific definition when considered in the context of a specific form of energy, work or power. The concept of efficiency, when applied in the thermodynamics domain, can involve power, energy or work. In thermodynamics, when power is the subject of analysis, efficiency is defined as follows: Efficiency = h = (Output Power)/(Input Power) Efficiency in percent = h = (Output Power)/(Input Power) × 100 Where, h (Eta) is a universal symbol for efficiency Also, in thermodynamics, when energy is the subject of analysis, efficiency is defined as follows: Efficiency = h = (Energy Put Out)/(Energy Taken In) Or, Efficiency = h = (Energy Put Out)/(Energy Taken In) × 100 Although work is not used as commonly in the computation of efficiency, in thermodynamics, where applicable, the efficiency calculation based on work would be as follows: Efficiency = h = (Work Performed by The System)/ (Work Performed on System)

34  Thermodynamics and Power Or, Efficiency = h = (Work Performed by The System)/ (Work Performed on System) × 100 As obvious from the definitions of efficiency above, since energy cannot be created, efficiency is always less than 1, or less than 100%. The decimal result for efficiency is often converted to, and stated as, a percentage value. In the following section, we will explore the relationship between power and efficiency in steam, mechanical and electrical systems, and develop better understanding of the flow of power in steam type electrical power generating systems. 2.2.8  Power – steam, mechanical and electrical The power delivered by steam to the turbine blades, Psteam, in a simplified - no heat loss, no kinetic head loss, no potential head loss and zero frictional head loss - scenario can be represented by the mathematical relationship stated in form of Eq. 2.1. In the context of flow of energy from steam to electricity, functional relationship between electrical power, PElectrical, generator efficiency ηGenerator, steam turbine efficiency ηTurbine, and Psteam can be expressed in form of Eq. 2.2. Psteam = (hi - hf)ṁ 

(2.1)

PElectrical = (Psteam)(ηTurbine)(ηGenerator)(2.2) The flow of power and energy from steam to electricity is depicted, in a power flow diagram, in Figure 2.1. This diagram is, essentially a pictorial illustration of the Eq. 2.2. The power flow diagram in Figure 2.1 also forms the crux of the scenario analyzed in Case Study 2.1, below.

2.3  Case Study 2.1. Steam to Electricity Conversion As an energy engineer, you are charged with the task to estimate the heat content or enthalpy, hi, of the superheated steam that must be fed to a steam turbine in order to supply 10 MW (Mega Watt) of electrical power to the electrical grid. Assume that there is no heat loss in the turbine system and that difference between the enthalpies on the entrance and exit ends of the turbine

2.3  Case Study 2.1. Steam to Electricity Conversion  35

Figure 2.1  Steam to Wire Power Flow – Steam Power Generation System.

is converted completely into work, minus the inefficiency of the turbine. All of the data available and pertinent to this project is listed below:

•• •• •• ••

Electrical Power Generator Efficiency: 90% Steam Turbine Efficiency: 70% Mass flow rate for steam, ṁ: 25 kg/s (55 lbm/s) Estimated exit enthalpy, hf, of the steam: 2875 kJ/kg (1239 BTU/lbm)

Solution Solution strategy In order to determine the estimated enthalpy, hi, of the incoming steam, we need to start with the stated output (10 MW) of the generator and work our way upstream to derive the energy delivered to the vanes of the turbine. The assumption that there is no heat loss in the turbine system and that the difference between the enthalpies on the entrance and exit ends of the turbine is

36  Thermodynamics and Power converted completely into work, minus the inefficiency of the turbine, implies that the energy delivered by the steam is equal to the net energy delivered to the turbine vanes. Also, note that net energy delivered to the turbine vanes is reduced or derated according to the given efficiency of the turbine. Solution in SI/Metric units Since, 1J/s = 1W and 1 kJ/s = 1kW, Power output of the generator = 10 MW = 10,000kW = 10,000kJ/s Brake horsepower delivered by the turbine to the generator, through the turbine shaft is determined as follows: BHP = Generator Output/Generator Efficiency = 10,000kJ/s /0.9 = 1.11 × 104 kJ/s or 11,111 kJ/s Power delivered by the steam to the turbine vanes is determined as follows: Psteam = BHP/Turbine Efficiency = (1.11 × 104 kJ/s)/0.7 = 1.5873 × 104 kJ/s or 15,873 kJ/s Of course, we could obtain the same result, in one step, by rearranging and applying Eq. 2.2 as follows: PElectrical = (Psteam )(ηTurbine) (ηGenerator)  Psteam = PElectrical /{(ηTurbine) (ηGenerator)} Psteam = (10,000 kJ/s)/ {(0.9)(0.7)} = 15,873 kJ/s

(2.2)

Since the difference in the turbine entrance and exit enthalpies, in this scenario, is equal to the energy delivered to the turbine vanes: Psteam = (hi - hf)ṁ(2.1) 15,873 kJ/s = (hi - 2875 kJ/kg)25 kg/s hi = (15,873 kJ/s)/(25 kg/s) + 2875 kJ/kg hi = 3,509 kJ/kg Solution in US/Imperial units Since, 1J/s = 1W and 1 kJ/s = 1kW,

2.3  Case Study 2.1. Steam to Electricity Conversion  37

Power output of the generator = 10 MW = 10,000kW = 10,000kJ/s Since 1.055 kJ = 1.0 BTU, Power output of the generator = (10,000kJ/s )(1/1.055kJ/BTU) = 9,479 BTU/s Brake horsepower delivered by the turbine to the generator, through the turbine shaft, is determined as follows: BHP = Generator Output/Generator Efficiency = (9,479 BTU/s) /0.9 = 10,532 BTU/s Power delivered by the steam to the turbine vanes is determined as follows: Psteam = BHP/Turbine Efficiency = (10,532 BTU/s)/0.7 = 15,046 BTU/s Alternatively, we could obtain the same result, in one step, by rearranging and applying Eq. 2.2 as follows: PElectrical = (Psteam )(ηTurbine) (ηGenerator)(2.2) Psteam = PElectrical /{(ηTurbine) (ηGenerator)} Psteam = (9,479 BTU/s)/{(0.9)(0.7)} = 15,046 BTU/s Since the difference in the turbine entrance and exit enthalpies, in this scenario, is equal to the energy delivered to the turbine vanes: Psteam = (hi - hf)ṁ(2.1) 15,046 BTU/s = (hi - 1239 BTU/lbm)(55 lbm/s) hi = (15,046 BTU/s)/( 55 lbm/s) + 1239 BTU/lbm hi = 1512 BTU/lbm

38  Thermodynamics and Power

Chapter 2 Self-Assessment Problems and Questions 1.

As an energy engineer, you are charged with the task to estimate the amount of electrical power produced, in MW, by a steam based power generating plant. Assume that there is no heat loss in the turbine system and that difference between the enthalpies on the entrance and exit ends of the turbine is completely converted into work, minus the inefficiency of the turbine. All of the data available, pertinent to this project, is listed below:

•• Electrical Power Generator Efficiency: 87% •• Steam Turbine Efficiency: 67% •• Mass flow rate for steam, ṁ: 20 kg/s (44 lbm/s) •• Exit enthalpy, h , of the steam: 2900 kJ/kg (1249 BTU/lbm) •• Incoming superheated steam enthalpy, h : 3586 kJ/kg (1545 BTU/ f

lbm)

2.

i

Consider the scenario described in Problem (1).Your client has informed you that the power generating plant output requirement has now doubled. Based on the concepts and principles learned in Chapter 2, what is the most suitable alternative for doubling the power output if the exit enthalpy, hf, of the steam must be kept constant at the original 2900 kJ/kg (1249 BTU/lbm) level?



a. Double the mass flow rate, ṁ, only.



b. Double the incoming superheated steam enthalpy, hi only.



c. Double the efficiency of the turbine.



d. Double the efficiency of the generator.



e. Increase mass flow rate, ṁ, incoming superheated steam enthalpy, hi and increase the efficiency specification on the turbine.

3 Study of Enthalpy and Entrophy

Topics ■■ ■■

Enthalpy Entropy

3.1 Introduction Like the last chapter, the goal in this brief chapter is to continue the introduction of basic, yet critical, concepts in the field of thermodynamics. In this chapter, we will introduce the concept of entropy and we will expand on the concept of enthalpy. As we progress through this text, you will notice that the discussion on entropy will be limited, reflecting the somewhat limited role of entropy in practical thermodynamics. On the other hand, our continued exploration of enthalpy, in this chapter, and the ones heretofore, is indicative of the instrumental and ubiquitous role of enthalpy in the study of thermodynamics. We received a brief, preliminary, introduction to enthalpy in the last chapter - in the context of energy flow in power generating realm. In this chapter, we will expand on enthalpy in preparation for its examination in more complex thermodynamic scenarios. 3.1.1 Enthalpy Enthalpy is defined as the total heat content or total useful energy of a substance. The symbol for enthalpy is “h.” Enthalpy is also considered to be the sum of internal energy “u” and flow energy (or flow work) p.V. This definition of enthalpy can be expressed, mathematically, premised on the law of conservation of energy, as follows: h = u + p.V 

(3.1)

Where, h = Specific enthalpy, measured in kJ/kg (SI Units) or BTU/lbm (US Units) 39

40  Study of Enthalpy and Entrophy u = Specific internal energy, measured in kJ/kg (SI Units) or BTU/lbm (US Units) p = Absolute Pressure measured in Pa (SI Units), or psf (US Units) V= Volume measured in m3 (SI Units), or ft3 (US Units) pV = Flow Energy, Flow Work or pV work, quantified in kJ/kg (SI Units) or BTU/lbm (US Units) In practical saturated or superheated steam systems, internal energy, u, specific enthalpy, h, and specific volume, u, can be assessed through saturated steam tables and superheated steam tables, respectively. The terms saturated steam and superheated steam are defined in depth later in this text. Chapters 5 and 6 cover classifications of steam and associated steam tables in detail. Reference steam tables, in US and SI form, are included in Appendix B of this text. In an effort to maintain consistency of units in practical thermodynamic situations, where computation is performed in US units, a more suitable form of the enthalpy equation Eq. 3.1would be as follows: h = u + pV/J 

(3.2)

Where, h = Enthalpy, measured in BTU’s u = Internal energy, measured in BTU p = Absolute Pressure measured in psf or lbf/ft2 V= Volume measured in ft3 J = Joule’s constant; value of J is 778 ft lbf/BTU Note that in SI unit system, an alternate version of enthalpy equation Eq. 3.1 is not necessary because units in Eq. 3.1 are congruent. Enthalpy can also be quantified in molar form. Molar units are often used by chemical engineers. In molar form, enthalpy is referred to as molar enthalpy and represented by the symbol “H”. The units for molar enthalpy H are BTU/lbmole, in the US system, and are kJ/kmole, in the Metric or SI System. Where a mole of a substance is defined or calculated through division of the mass of that substance by the atomic weight of the substance, if it is a solid, or by the molecular weight, if it is a liquid or gas. The mathematical equation for molar enthalpy “H,” is as follows: H = U + pV 

(3.3)

Where, U = Molar Internal Energy, can be expressed in BTU/lbmol (US Units) or kJ/kmol (SI Units)

3.1 Introduction  41

p = Absolute pressure measured in Pa (SI Units), psf (US Units) or lbf/ft2 V= Molar specific volume measured in m3/kmol (SI Units), or ft3/ lbmole (US Units) Example 3.1 Calculate the absolute enthalpy, h, in BTU’s, for 1 lbm of vapor under the following conditions: h = Enthalpy, measured in BTU’s =? u = 1079.9 BTU/lbm p = 14.14 psia V = 27.796 ft3 J = Joule’s constant; value of J is 778 ft lbf/BTU Solution The pressure is given in psia, or lbf/in2. In order to streamline the pressure for application in Eq. 3.2, we must convert in into lbf/ft2. Therefore, p = (14.14 lbf/in2)(144 in2/ ft2) = 2,036 lbf/ft2 Then, by applying Eq. 3.2, and by substitution of known and derived values: h = u + pV/J (3.2) h = 1079.9 BTU/lbm + (2,036 lbf/ft2)(27.796 ft3 )/ 778 ft lbf/BTU h = 1152.67 BTU 3.1.2 Entropy Entropy is defined as the non-work producing form of energy. It is also regarded as the energy that is not available for performing useful work within a certain environment. The symbol for entropy is “s.” Some facts, principles and laws associated with entropy are summarized below:

•• ••

Increase in entropy is referred to as entropy production. The total absolute entropy of a system is said to be equal to the sum of all absolute entropies that have occurred over the life of the system. stotal = ∑ Δsi 

(3.4)

42  Study of Enthalpy and Entrophy Where, Δsi represents change in enthalpy at each object or in each substance. Application of this entropy principle will be demonstrated through Case Study 3.1. According to the third law of thermodynamics, the absolute entropy of a perfect crystalline solid, in thermodynamic equilibrium, approaches zero as the temperature approaches absolute zero. T Limit → 0° K s → 0 In an isothermal (constant temperature) process, the entropy production, Δs, is a function of the energy transfer rate: Δs = q / Tabs(3.5) Where, s q T abs

= entropy in kJ/kg.°K (SI Units System), or in BTU/lbm.°R (US Unit System) = Heat transferred in kJ/kg, (SI Units) or BTU/lbm (US Units) = Absolute Temperature of the object or substance, in°K (SI Units System), or in°R (US Unit System)

3.2  Case Study 3.1: Entropy Analysis In a certain solar system, there are four (4) planets oriented in space as shown in Figure #2. Their temperatures are indicated in the diagram, in°K as well as in°R. As apparent from the orientation of these planets in Figure 3.1, they are exposed to each other such that heat transfer can occur freely through radiation. All four (4) planets are assumed to be massive enough to allow for the interplanetary heat transfer to be isothermal for each of the planets. a. Will heat transfer occur, through radiation, from planet Z to planets × and Y? b. If the 3,000 kJ/kg of radiated heat transfer occurs from planet × to planet Y, what would be the entropy changes at each of the two planets? c. Can convectional heat transfer occur between any of two planets in this solar system? d. If certain radiated heat transfer between Planets Y and Z causes an entropy change of 11.77 kJ/kg.°K at Planet Y and an entropy change

3.2  Case Study 3.1: Entropy Analysis  43

Figure 3.1  Case Study 3.1, Entropy.

of 12.66 kJ/kg °K at Planet Z, what would be the overall, resultant, entropy of this planetary system? e. Can planet × be restored to its original state? If so, how? Solution - Case Study 3.1 Will heat transfer occur, through radiation, from planet Z to planets × and Y? Solution/Answer Heat flows from a body at higher temperature to one that is at lower temperature. The temperature of Planet Z is lower than the temperature of planets × and Y. Therefore, NO radiated heat transfer will occur from planet Z to planets × and Y. b) If the 3000 kJ /kg of radiated heat transfer occurs from planet × to planet Y, what would be the entropy changes at each of the two planets? Solution/Answer: In an isothermal (constant temperature) process, the entropy production, Δs, is a function of the energy transfer rate and its relationship with heat q and absolute temperature, T abs is represented by Eq. 3.5: Δs = q / T abs  ΔsX = (- 3,000 kJ/kg)/(290°K) = - 10.34 kJ/kg °K {Due to heat loss by Planet X}

(3.5)

44  Study of Enthalpy and Entrophy And,

ΔsY = (+3,000 kJ/kg)/(280°K) = + 10.71 kJ/kg °K {Due to heat gain by Planet Y}

c) Can convectional heat transfer occur between any of two planets in this solar system? Solution/Answer Convectional heat transfer is dependent on bulk movement of a fluid (gaseous or liquid) and, therefore, it can only occur in liquids, gases, and multiphase mixtures. Since the system in this problem is a planetary system, the medium between the bodies is devoid of fluids or medium needed for convectional heat transfer. Heat transfer in this planetary system occurs through radiation, primarily. Therefore, the answer is NO. d) If the heat transfer from part (b) occurs simultaneous to a certain radiated heat transfer between Planets Y and Z, - where the entropy change of - 11.77 kJ/kg °K is recorded at Planet Y and an entropy change of12.66 kJ/kg °K is recorded at Planet Z - what would be the overall, resultant, entropy of this planetary system? Solution/Answer Overall ΔsPlanetary System = ∑(Δsi ) ∴ Overall ΔsPlanetary System = ΔsX + ΔsY + ΔsYZ + ΔsZ Or,

 10.34 kJ/kg °K + 10.71 kJ/kg °K - 11.77 kJ/kg °K Δs Planetary System = + 12.66 kJ/kg °K ∴ Overall Δs Planetary System = + 1.2643 kJ/kg °K

e) Can planet × be restored to its original state? If so, how? Solution/Answer Planet × can be restored to its original state; through absorption of +3,000 kJ/kg of (specific) heat energy.

Self-Assessment Problems and Questions  45

Chapter 3 Self-Assessment Problems and Questions 1.

Calculate the volume 1 kg of vapor would occupy under the following conditions: h = 2734 kJ u = 2550 kJ p = 365.64 kPa = 365.64 kN/m2 V=?

2.

In a certain solar system there are four (4) planets oriented in space as shown in Figure 3.1. As apparent from the orientation of these planets, they are exposed to each other such that heat transfer can occur freely through radiation. All four (4) planets are assumed to be massive enough to allow for the interplanetary heat transfer to be an isothermal phenomenon for each of the planets. Perform all computation in the US Unit System.



a. If the 1,300 BTU/lbm of radiated heat transfer occurs from planet × to planet Y, what would be the entropy changes at each of the two planets?



b. If a certain radiated heat transfer between Planets Y and Z causes an entropy change of -2.9 BTU/lbm.°R at Planet Y and an entropy change of 3.1 BTU/lbm.°R at Planet Z, what would be the overall, resultant, entropy of this planetary system?

3.

If the mass of vapor under consideration in problem 1 were tripled to 3 kg, what would be the impact of such a change on the volume?

4.

Would Eq. 3.2 be suitable for calculation of enthalpy if all available data is in SI (Metric) units?

4 Understanding Mollier Diagram

Topic ■■

Mollier diagram, its relationship with the psychrometric chart, and its applications in thermodynamics.

4.1 Introduction Mollier diagram is named after Richard Mollier (1863–1935), a German professor who pioneered experimental research on thermodynamics associated with water, steam and water-vapor mixture. Mollier diagram is a graphical representation of functional relationship between enthalpy, entropy, temperature, pressure and quality of steam. Mollier is often referred to as Enthalpy – Entropy Diagram or Enthalpy – Entropy Chart. It is used commonly in the design and analysis associated with power plants, steam turbines, compressors, and refrigeration systems. Mollier diagram is available in two basic versions: The SI/Metric unit version and the US/Imperial unit version. Figure 4.1 depicts an SI/Metric version of the Mollier diagram. The US Customary version of Mollier diagram is depicted in Figure 4.2. In the SI version of the Mollier Diagram below, the abscissa (horizontal or x-axis in a Cartesian coordinate system) and ordinate (vertical or y-axis in a Cartesian coordinate system) scales represent entropy and enthalpy, respectively. Hence, Mollier diagram is also referred to as the Enthalpy-Entropy Chart. The constant pressure and constant temperature lines in the Mollier diagram are referred to as isobars and isotherms, respectively. In addition, the graph includes lines representing constant steam quality, or quality factor, “x,” in the bottom half of the diagram. The bold line, spanning from left to right, in the lower half of Mollier diagram is the saturation line. The saturation line, labeled as x = 1, represents the set of points on Mollier diagram where the steam is 100% vapor, by mass. All points above the saturation line are in the superheated steam realm. All points below the saturation line 47

48  Understanding Mollier Diagram

Figure 4.1  Mollier diagram, SI/Metric units.

represent a mixture of liquid and vapor phases – with the quality factor or “x” lines representing, specific, quality factors in “%.” The segment of the Mollier diagram that lies below the saturation line (x = 1), is the region where typical saturated steam tables apply. The concept of quality or quality factor is explained and illustrated in Chapter 5. As noted in the US unit

4.1 Introduction  49

Figure 4.2  Mollier diagram, US customary units.

version of the Mollier diagram below, the quality factor “x” is, sometimes, referred to as the “steam quality,” also expressed in “%.” One school of thought proclaims the Mollier diagram as the European version of the Anglo-American psychrometric chart. They are identical in content but not in appearance.

50  Understanding Mollier Diagram

Figure 4.3  Mollier diagram transformation to psychrometric chart.

A comparison of the Mollier diagram and the psychrometric chart reveals convincing similarity between these two versatile and commonly applied thermodynamics tools. Some schools of thought explain the process of transformation of the Mollier diagram to the psychrometric chart on the basis of geometric manipulation. This relationship between Mollier diagram and the psychrometric chart is apparent from the fact that both involve critical thermodynamic properties such as enthalpy, temperature, sensible heat, latent heat and quality. The transformation, or perhaps evolution, of Mollier Diagram to Psychrometric Chart is depicted in Figure 4.3 below. In figure 4.3, above, “i” represents the heat content, enthalpy, or specific enthalpy, expressed in kJ/Kg or BTU/lbm, denoted elsewhere in this text as “h.” The symbol “t” denotes temperature, specifically, in this case, the dry

4.2  Application of Mollier Diagram  51

bulb temperature. The temperature would be expressed in ˚F or ˚C, in US Customary Units and SI Units, respectively. The symbol “x” denotes vapor quality factor, expressed in decimal form or “%.” The quality factor bears direct relationship to relative humidity. The Mollier diagram is also referred to as the “ix” diagram based on the relationship between heat content, i and water vapor content of air, x. The heat, or energy, content is difficult to measure directly, so the diagram is “cleverly” contorted to give the illusion of being based on the relationship between temperature, relative humidity, and water vapor content. Temperature and relative humidity are, relatively, easy to measure, so Mollier diagram serves as a versatile tool for analyzing and designing steam and HVAC systems. A comparison of the Mollier diagram and the steam tables also reveals a marked similarity and equivalence between the two. This equivalence is illustrated through Example 5.6 in Chapter 5. The reader would be better prepared to appreciate the illustration of relationship between the Mollier diagram and the steam tables after gaining a clear comprehension of the saturated and superheated steam tables in Chapter 5. Example 5.6 demonstrates the interchangeability of the Mollier diagram and the Superheated Steam Tables as equivalent tools in deriving the enthalpy values associated with the change in the temperature of superheated steam. This equivalence between Mollier diagram and the steam tables is further reinforced by the fact that both involve critical thermodynamic properties of steam such as enthalpy, entropy, temperature and pressure.

4.2  Application of Mollier Diagram A common application of Mollier diagram involves determination of an unknown parameter among the key Mollier diagram parameters such as, enthalpy, entropy, temperature pressure and quality. Typical applications of Mollier diagram are illustrated through the example problems that follow. Example 4.1 Determine if steam at 450°C and 1 bar is saturated or superheated. Find the enthalpy and entropy of this steam. Solution See the Mollier diagram in Figure 4.4. Identify the point of intersection of the 450°C line (or 450°C isotherm) and the constant pressure line (or isobar) of 1 bar. This point of intersection of the two lines is labeled A.

52  Understanding Mollier Diagram

Figure 4.4  Mollier diagram, SI/Metric units.

4.2  Application of Mollier Diagram  53

As explained above, this region of the Mollier diagram is the superheated steam region. Therefore, the steam at 450°C and 1 bar is superheated. 4.2.1  Enthalpy determination To determine the enthalpy at point A, draw a straight horizontal line from point A to the left till it intersects with the diagonal enthalpy line. This horizontal line intersects the enthalpy line at an enthalpy value of, approximately, 3380 kJ/kg. Therefore, hA, or enthalpy at point A, is 3380 kJ/kg. 4.2.2  Entropy determination To determine the entropy at point A, draw a straight vertical line from point A to the bottom, until it intersects with the entropy line. The vertical line intersects the entropy line at, approximately, 8.7kJ/kg.°K. Therefore, sA, or entropy at point A, is 8.7kJ/kg °K. Example 4.2 Determine the amount of heat that must be removed from a system, on per kg basis, to reduce the temperature of steam from 450°C, at 1 Atm. to 400°C, at 1 Atm. Solution To determine the amount of heat that must be removed from the steam in order to cool the steam from 450°C, at 1 Atm. to 400°C, at 1 Atm, we must assess the enthalpies at those two points. The first point, at 450°C and 1 Atm, was labeled as point A in Example 4.1. The enthalpy, hi at point A was determined to be 3380 kJ/kg. The enthalpy, hf, at the second point – referred to as point B – as shown on the Mollier diagram in Figure 4.2, is 3280 kJ/kg. Therefore, the amount of heat that must be removed from the system in order to lower the temperature from 450°C to 400°C, at 1 Atm, would be: Δh = hi - hf \ Δh = hi - hf = 3380 kJ/kg - 3280 kJ/kg = 100 kJ/kg In other words, 100 kJ of heat must be removed from each kg of steam in order to cool it from 450°C, at 1 Atm. to 400°C, at 1 Atm.

54  Understanding Mollier Diagram

Chapter 4 Self-Assessment Problems and Questions 1.

Using the Mollier diagram, find the entropy of steam at 400°C and 1 Atm.

2.

Heat is removed from a thermodynamic system such that the temperature drops from 450°C, at 1 Atm to 150°C, at 1 Atm. Determine the following:



a. The new, or final, Enthalpy



b. The new entropy



c. The state of steam at 150°C and 1 Atm

5 Saturated and Superheated Steam Tables

Topics ■■ ■■

Saturated Steam Tables Superheated Steam Tables.

5.1 Introduction In this text, as we study various topics of thermodynamics, we will utilize and focus on two main categories of steam tables: (1) The Saturated Steam Tables and (2) The Superheated Steam Tables. Appendix B of this text includes the compact version of the saturated steam tables and the superheated steam tables. These tables are referred to as the compact version because they do not include certain properties or attributes that are customarily included only in the detailed or comprehensive version. Characteristics or properties included in most comprehensive version of the saturated steam tables, but omitted in Appendix B steam tables, are as follows: 1.

Internal energy “U.”

2.

The heat of vaporization “hfg”

3.

Duplicate pressure of temperature columns - typically included in the comprehensive steam tables to avoid parallax error while retrieving data from the table.

Internal energy, absolute and specific, is not required in most common thermodynamic analysis. And, heat of vaporization, hfg for water - as explained in Chapter 6 – is a derivative entity. In that, hfg can be derived from hL and hv as stipulated by Eq. 5.1 below: hfg = hv - hL(5.1) 55

56  Saturated and Superheated Steam Tables Example 5.1 Using the saturated liquid enthalpy value for hL and the saturated vapor enthalpy value for hV, at 1 MPa and 180°C, as listed in the saturated steam table excerpt in Table 5.1, verify that hfg = 2015 kJ/kg. Solution As stated in Eq. 5.1: hfg = hV - hL As read from Table 5.1: hV hL

= 2777 kJ/kg, and = 762.68 kJ/kg

\ hfg = hV - hL = 2777 - 762.68 = 2014.32 kJ/kg The value for hfg, at 1.0 MPa and 180°C, as listed in Table 5.1, is 2015 kJ/kg, versus the derived value of 2014 kJ/kg. The difference between the calculated value of hfg, at 1.0 MPa and 180°C and the value listed in Table 5.1 is only 0.05% and is, therefore, negligible. Hence, we can say that Eq. 5.1 stands verified as a tool or method for deriving the heat of vaporization hfg from the compact version of steam tables included in Appendix B. The saturated and superheated steam tables in Appendix B are presented in the US/Imperial unit realm as well as the SI/Metric realm. Note that in this chapter - as well as other chapters in this text - for the readers’ convenience, saturated steam table excerpts include the heat of vaporization, hfg, values. See Tables 5.1, 5.3, 5.4, and 5.5. Also, for illustration of various numerical examples, and thermodynamics discussion in general, excerpts from the superheated steam tables in Appendix B, are included in this chapter in form of Tables 5.2, 5.6 and 5.7.

5.2  Saturated Steam Tables Saturated water and steam tables, as presented in Appendix B, are categorized as follows: A. Saturated water and steam tables, by temperature, in US Units

Table 5.1  Properties of saturated steam, by pressure, SI units.

Properties of Saturated Steam by Pressure Metric/SI Unitss Specific Volume m3/kg

Enthalpy kJ/kg

Entropy kJ/kg

Abs. Press. MPa

Temp°C

0.010

45.81

0.0010103

14.671

191.81

2392.8

2583.9

0.6492

8.1489

0.010

0.10

99.61

0.0010431

1.6940

417.44

2258.0

2674.9

1.3026

7.3588

0.10

0.20

120.21

0.0010.605

0.88574

504.68

2201.9

2706.2

1.5301

7.1269

0.20

1.00

179.89

0.0011272

0.19435

762.68

2015.3

2777.1

2.1384

6.5850

1.0

Sat. Liquid Sat. Vapor Sat. Liquid VL VV hL

Evap. hfg

Abs. Sat. Vapor Sat. Liquid Sat. Vapor Press. hV SL SV MPa

5.2  Saturated Steam Tables  57

58  Saturated and Superheated Steam Tables B.

Saturated water and steam tables, by pressure, in US Units

C.

Saturated water and steam tables, by temperature, in SI/Metric Units

D. Saturated water and steam tables, by pressure, in SI/Metric Units A. Saturated water and steam tables, by temperature, in US Units As apparent from the inspection of the four categories of saturated steam tables above, two distinguishing factors between these categories of tables are temperature and pressure. First category of tables, listed under bullet A, represents saturated water and steam data by temperature, in US Units. In other words, this set of tables is used when temperature is the determining factor, or when the current or future state of the saturated water or saturated steam is premised on or defined by the temperature. So, if saturated water or saturated steam is said to exist at a given temperature, the following properties can be identified: a. Saturation pressure, in psia, at the given temperature, in°F. b. Specific volume, nL, in ft3/lbm, of saturated liquid, at the given temperature and saturation pressure. c. Specific volume, nv, in ft3/lbm, of saturated vapor, at the given temperature and saturation pressure. d. Specific enthalpy, hL, in BTU/lbm, of saturated liquid, at the given temperature and saturation pressure. e. Specific enthalpy, hv, in BTU/lbm, of saturated vapor, at the given temperature and saturation pressure. f. Specific entropy, sL, in BTU/lbm °R, of saturated liquid, at the given temperature and saturation pressure. g. Specific entropy, sv, in BTU/lbm °R, of saturated vapor, at the given temperature and saturation pressure. B. Saturated water and steam tables, by pressure, in US units The second category of tables represents saturated water and steam data by pressure, in US Units. In other words, this set of tables is used when pressure is the determining factor, or when the current or future state of the saturated water or saturated steam is defined by the pressure. So, if saturated water or

5.3  Superheated Steam Tables  59

saturated steam is said to exist at a given pressure, the following properties can be identified: a. Saturation temperature, in °F, at the given pressure, in psia °F. b. Specific volume, nL, in ft3/lbm, of saturated liquid, at the given pressure and saturation temperature. c. Specific volume, nv, in ft3/lbm, of saturated vapor, at the given pressure and saturation temperature. d. Specific enthalpy, hL, in BTU/lbm, of saturated liquid, at the given pressure and saturation temperature. e. Specific enthalpy, hv, in BTU/lbm, of saturated vapor, at the given pressure and saturation temperature. f. Specific entropy, sL, in BTU/lbm °R, of saturated liquid, at the given pressure and saturation temperature. g. Specific entropy, sv, in BTU/lbm °R, of saturated vapor, at the given pressure and saturation temperature. C&D Saturated steam tables categorized as C and D above are very similar to categories A and B, with the exception of the fact that the temperature, pressure, specific volume, enthalpy and entropy are in the metric unit system.

5.3  Superheated Steam Tables Superheated steam tables, as presented in Appendix B, are categorized as follows: A. Superheated steam tables in US units B.

Superheated steam tables in SI/Metric units

Unlike the saturated steam tables, regardless of the unit system, the superheated steam tables differ from the saturated steam tables as follows: a. Superheated steam tables, such as the ones included under Appendix B, provide only the specific volume, specific enthalpy, and specific entropy, for a given set of temperature and pressure conditions. b. Retrieval of specific values of enthalpy and entropy from the superheated steam tables requires knowledge of the exact temperature and pressure.

60  Saturated and Superheated Steam Tables c. When the exact temperature and pressure for a given superheated steam condition are not available or listed in the superheated steam tables, single or double interpolation is required to identify the specific volume, enthalpy and entropy.

5.4  Single and Double Interpolation of Steam Table Data Interpolation is often required when the retrieving data from tables such as the Saturated Steam Tables or the Superheated Steam Tables. Interpolation is needed when the given pressure or temperature don’t coincide with the exact standard pressure and temperature values on the given tables. Example 5.2 offers an opportunity to study the interpolation method, in the US unit realm. Even though the interpolation method is being illustrated on the basis of steam tables in this chapter, this technique can be employed for interpolation of other types of tabular data, as well. Example 5.2 Calculate the enthalpy of 450 psia and 950°F superheated steam. Solution As you examine the superheated steam tables for these parameters, in Appendix B, you realize that exact match for this data is not available in the table. See Tables 5.2 and 5.3, below, for excerpts from the superheated steam tables in Appendix B. While the given pressure of 450 psia is listed, the stated temperature of 950°F is not listed. Therefore, the enthalpy for 450 psia and 950°F superheated steam and must be derived by applying interpolation to the enthalpy data listed in the tables for 900°F and 1,000°F. The formula for single interpolation, applied between the stated or available enthalpy values for 900°F and 1000°F, at 450 psia, is as follows: h 950°F, 450 psia = ((h1000°F, 450 psia - h900°F, 450 psia )/(1000°F -900°F))(950-900) + h 900°F, 450 psia By substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 5.2: h 950°F, 450 psia = ((1522.4 BTU/lbm – 1468.6 BTU/lbm )/ (1000°F - 900°F))(950-900) + 1468.6 BTU/lbm = 1496 BTU/lbm

5.4  Single and Double Interpolation of Steam Table Data  61 Table 5.2  Superheated steam table excerpt, US/Imperial units.

Properties of Superheated Steam US/Imperial Units Abs. Press. psia

Temp. °F 

(Sat. Temp. °F )

Note: n is in ft3/lbm, h is in BTU/lbm and s is in BTU/(lbm-°R)

500

700

900

1000

n

2.062

2.5818

3.0683

3.3065

h

1262.5

1370.8

1475.2

1527.8

 

s

1.5901

1.6928

1.7758

1.8132

360 (434.43)

n

1.446

1.8429

2.2028

2.3774

h

1250.6

1365.2

1471.7

1525

 

s

1.5446

1.6533

1.7381

1.7758

450 (456.32)

n

1.1232

1.4584

1.7526

1.8942

h

1238.9

1360

1468.6

1522.4

 

s

1.5103

1.6253

1.7117

1.7499

600 (486.25)

n

 

 

1.3023

1.411

h

 

 

1463.2

1518

s

 

 

1.577

1.7159

260 (404.45)

 

Note: The available enthalpy values are circled in Table 5.2. Example 5.3 offers an opportunity to study the double interpolation method. As is the case with single interpolation method, even though the double interpolation method is being illustrated on the basis of steam tables in this chapter, this technique can be employed for double interpolation of other types of tabular data, as well.

62  Saturated and Superheated Steam Tables Example 5.3 Calculate the enthalpy of 405 psia and 950°F superheated steam. Solution: As you examine the superheated steam tables for these parameters, in Appendix B, you realize that exact match for this data is not available. See Table 5.2 for an excerpt of the superheated steam tables in Appendix B. In this example, neither the given pressure of 405 psia nor the stated temperature of 950°F is listed in Appendix B superheated steam tables. Therefore, the enthalpy for 405 psia and 950°F superheated steam and must be derived by applying double interpolation to the enthalpy data listed in the Table 5.2 for 360 psia1, 450 psia, 900°F and 1,000°F. Note: Since the enthalpy data for 400 psia is available in Appendix B, double interpolation could be performed on 400 psia and 450 psia points yielding the same results. The lower pressure point of 360 psia is chosen in this example simply to maintain continuity with the superheated steam table excerpt in Table 5.2.

1

The double interpolation approach, as applied here, will entail three steps. The first step involves determination of h 900°F, 405 psia, the enthalpy value at 405 psia and 900°F. The enthalpy values available and used in this first interpolation step are circled in Table 5.2. The following formula sums up the mathematical approach to this first step: h 900°F, 405 psia = ((h 900°F, 360 psia - h 900°F, 450 psia )/(450 psia -360 psia)) (450 psia - 405 psia) + h 900°F, 450 psia Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 5.2: h 900°F, 405 psia = ((1471.7 BTU/lbm – 1468.6 BTU/lbm)/(450 psia -360 psia)) (450 psia - 405 psia) + 1468.6 BTU/lbm = 1470 BTU/lbm Second interpolation step involves determination of h 1000°F, 405 psia, the enthalpy value at 405 psia and 1000°F. The enthalpy values available and used in this

5.4  Single and Double Interpolation of Steam Table Data  63

interpolation step are circled in Table 5.2. The following formula sums up the mathematical approach associated with this interpolation step:

h 1000°F, 405 psia = ((h 1000°F, 360 psia - h 1000°F, 450 psia )/(450 psia -360 psia)) (450 psia - 405 psia) + h 1000°F, 450 psia

Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 5.2: h 1000°F, 405 psia = ((1525 BTU/lbm – 1522.4 BTU/lbm )/(450 psia -360 psia)) (450 psia - 405 psia) + 1522.4 BTU/lbm = 1524 BTU/lbm The final step in the double interpolation process, as applied in this case, involves interpolating between h 1000°F, 405 psia and h 900°F, 405 psia, the enthalpy values derived in the first two steps above, to obtain the desired final enthalpy h 950°F, 405 psia. The formula for this final step is as follows: h 950°F, 405 psia = ((h 1000°F, 405 psia – h 900°F, 405 psia)/( 1000°F - 900°F))( 950°F - 900°F) + h 900°F, 405 psia Substituting enthalpy values derived in the first two steps above: h 950°F, 405 psia = ((1524 BTU/lbm – 1470 BTU/lbm)/( 1000°F - 900°F)) (950°F - 900°F) + 1470 BTU/lbm = 1497 BTU/lbm Example 5.4 Determine the enthalpy of saturated water at 20°C and 1 Bar. Solution The saturation temperature at 1 Bar, 1 Atm, or 101 kPa, as stated in the saturated steam tables in Appendix B, is 99.6°C or, approximately, 100°C. The saturated water in this problem is at 20°C; well below the saturation temperature. Therefore, the water is in a subcooled state.

64  Saturated and Superheated Steam Tables In the subcooled state, saturated water’s enthalpy is determined by its temperature and not the pressure. Hence, the enthalpy of saturated water at 20°C must be retrieved from the temperature based saturated steam tables. From Appendix B, and as circled in Table 5.3, the enthalpy of saturated water at 20°C is 83.92 kJ/kg 2. Note: Since the water is referred to as “saturated water” and is clearly identified to be subcooled, the enthalpy value selected from the tables is hL and not hV.

2

5.5 Quality of Steam Consideration in Thermodynamic Calculations In thermodynamics, there are myriad scenarios where water exists, simultaneously, in liquid and vapor forms. In such conditions, the concept of quality of steam plays a vital role. Quality, as described earlier is the ratio of the mass of vapor and the total mass of vapor and liquid. Mathematically, quality is defined as follows: Quality = x = Where, x mvapor mliquid mliquid + mvapor

mvapor mvapor + mliquid = Quality, or quality factor = mass of vapor in the liquid and vapor mixture = mass of liquid in the liquid and vapor mixture = Total mass of the liquid and vapor mixture

When quality of steam is less than one (1), or less than100%, determination of enthalpy - and other parameters that define the state of water under those conditions – requires consideration of the proportionate amounts of saturated water (liquid) and vapor. For instance, if the quality of steam is 50%, determination of total enthalpy would entail 50% of the enthalpy contribution from saturated vapor and 50% from saturated water (liquid). This principle is formulated mathematically through equations 5.2, 5.3, 5.4 and 5.5, and illustrated through the Example 5.5. The basic formulae for computing enthalpy, entropy, internal energy and specific volume when quality of steam is less than 100% are as follows: hx = (1-x)hL + xhV

(5.2)

Properties of Saturated Steam by Temperature Metric/SI Unitss Specific Volume m3/kg

Enthalpy kJ/kg

Entropy kJ/kg

Temp. °C

Abs. Press. Sat. Liquid Sat. Vapor Sat. Liquid VL VV hL MPa

Evap. hfg

Sat. Vapor Sat. Liquid Sat. Vapor Temp. hV SL SV °C

20

0.002339

0.0010018

57.7610

83.920

2454.1

2537.5

0.2965

8.6661

20

50

0.012351

0.0010121

12.0280

209.34

2382.7

2591.3

0.7038

8.0749

50

100

0.101420

0.0010435

1.6719

419.10

2257.0

2675.6

1.3070

7.3541

100

200

1.554700

0.0011565

0.1272

852.39

1940.7

2792.1

2.3308

6.4303

200

5.5  Quality of Steam Consideration in Thermodynamic Calculations  65

Table 5.3  Properties of saturated steam, by temperature, SI units.

66  Saturated and Superheated Steam Tables sx = (1-x)sL + xsv(5.3) ux = (1-x)uL + xuv(5.4) nx = (1-x)nL + xnv(5.5) Example 5.5 Determine the enthalpy and specific volume for 100 psia steam with a quality of 55%. Solution Given: Quality, × = 0.55 Absolute Pressure = 100 psia From saturated steam tables in Appendix B, and the excerpt in Table 5.4, the values of enthalpies and specific volumes, at 100 psia, are: hL = 298.57 BTU/lbm hV = 1187.5 BTU/lbm nL = 0.017736 ft3/lbm nv = 4.4324 ft3/lbm Apply equations 5.2 and 5.5: hx = (1-x)hL + xhV(5.2) nx = (1-x)nL + xnv(5.5) Then, hx = (1- 0.55)(298.57 BTU/lbm) + (0.55)(1187.5 BTU/lbm) hx = 787.48 BTU/lbm And,

nx = (1- 0.55)(0.017736) + (0.55)(4.4324 ft3/lbm) nx = 2.446 ft3/lbm

Properties of Saturated Steam by Pressure US/Imperial Units Specific Volume ft3/lbm

Enthalpy Btu/lbm

Entropy btu/(lbm.°R)

Abs. Press. psia

Temp. °F

1.0

101.69

0.016137

333.51

69.728

1036

1105.4

0.1326

1.9776

1.0

40.

152.91

0.016356

90.628

120.89

1006.4

1126.9

0.2198

1.8621

4.0

14.0

209.52

0.016697

28.048

177.68

972.0

1149.4

0.3084

1.7605

14.0

100

327.82

0.017736

4.4324

298.57

889.2

1187.5

0.4744

1.6032

100

Sat. Liquid Sat. Vapor Sat. Liquid VL VV hL

Evap. hfg

Abs. Sat. Vapor Sat. Liquid Sat. Vapor Press. hV SL SV psia

5.5  Quality of Steam Consideration in Thermodynamic Calculations  67

Table 5.4  Properties of saturated steam, by pressure, US units.

Properties of Saturated Steam by Temperature US/Imperial Units Specific Volume ft3/lbm

Enthalpy Btu/lbm

Entropy btu/(lbm.°R)

Temp. °F

Abs. Press. Sat. Liquid Sat. Vapor Sat. Liquid VL VV hL psia

Evap. hfg

Sat. Vapor Sat. Liquid Sat. Vapor Temp. hV SL SV °F

50

0.17813

0.016024

1702.9

18.066

1065.2

1083.1

0.0361

2.1257

50

100

0.95044

0.016131

349.87

68.037

1037.0

1104.7

0.1296

1.9819

100

210

14.1360

0.016701

27.796

178.17

971.6

1149.5

0.3092

1.7597

210

250

29.8430

0.017001

13.816

218.62

945.6

1164.0

0.3678

1.700

250

68  Saturated and Superheated Steam Tables

Table 5.5  Properties of saturated steam, by temperature, US units.

5.5  Quality of Steam Consideration in Thermodynamic Calculations  69 Table 5.6  Properties of superheated steam, SI units.

Properties of Superheated Steam Metric/SI Units Abs. Press. MPa (Sat. T, °C )   0.05 (81.33)     0.1 (99.61)     1.0 (179.89)     2.5 (223.99)     3.0 (233.86)     4.0 (250.36)  

  n h s   n h s   n h s   n h s   n h s   n h s

150   3.889 2780.1 7.9401   1.9367 2776.6 7.6147                                

300   5.284 3075.5 8.5373   2.6398 3074.5 8.2171   0.2580 3051.7 7.1247   0.0989 3008.8 6.6438   0.0812 2994.3 6.5412   0.0589 2961.7 6.3638

Temp. °C 500   7.134 3488.7 9.1546   3.5656 3488.1 8.8361   0.3541 3479.0 7.7640   0.13998 3462.1 7.3234   0.1162 3457.0 7.2356   0.0864 3445.8 7.0919

650                   0.4245 3810.5 8.1557   0.1623 3799.7 7.7056   0.1405 3797.0 7.6373   0.1049 3790.2 7.4989

800                   0.4944 4156.2 8.5024   0.1896 4148.9 8.0559   0.1642 4147.0 7.9885   0.1229 4142.5 7.8523

70  Saturated and Superheated Steam Tables Table 5.7  Properties of superheated steam, US units. n = specific volume in ft3/lbm; h = enthalpy in BTU/lbm; s = entropy in BTU/lbm-°R.

Abs. Press. psia (Sat. T, °F )   10 (193.16)     15 (212.99)     100 (327.82)     200 (381.81)     360 (434.43)  

Properties of Superheated Steam US/Imperial Units Temp. °F    n h s   n h s   n h s   n h s   n h s

200   38.851 1146.4 1.7926  

300          

500          

1000          

1500          

       

29.906 1192.7 1.8137  

37.986 1287.3 1.9243  

57.931 1534.7 2.1312  

       

       

       

5.5875 1279.3 1.7089  

8.6576 1532.3 1.9209  

       

       

       

2.7246 1269.1 1.6243  

4.3098 1529.5 1.8430  

       

     

     

1.4460 1250.6 1.5446

2.3774 1525.0 1.7758

3.2291 1799.8 1.9375

5.5  Quality of Steam Consideration in Thermodynamic Calculations  71

Example 5.6 Prove the equivalence of the Mollier Diagram and the Steam Tables by verifying the results of Example 4.1, Chapter 4, through the use of Steam Tables in Appendix B. Solution The solution from Example 4.1, as restated with the aid of Figure 5.1, is as follows: hi = Enthalpy at 450°C and 1 Atm., as read from the Mollier Diagram = 3380 kJ/kg hf = Enthalpy at 400°C and 1 Atm., as read from the Mollier Diagram = 3280 kJ/kg Then, using the Mollier diagram, the amount of heat that must be removed from the system in order to lower the temperature from 450°C to 400°C, at 1 Atm, would be: Δh = hi - hf = 3380 kJ/kg - 3280 kJ/kg = 100 kJ/kg Now, let’s determine the amount of heat to be removed using the steam tables, in Appendix B: hi = Enthalpy at 450°C and 1 Atm., from Appendix B = 3382.8 kJ/kg hf = Enthalpy at 400°C and 1 Atm., from Appendix B = 3278.5 kJ/kg So, using the Steam Tables, the amount of heat that must be removed from the system in order to lower the temperature from 450°C to 400°C, at 1 Atm, would be: Δh = hi - hf = 3382.8 kJ/kg - 3278.5 kJ/kg = 104.3 kJ/kg Therefore, for most practical purposes, the Mollier diagram and the Steam Tables are equivalent insofar as thermodynamic system analyses are concerned. The 4.3% difference between the two approaches is due mainly to the small amount of inaccuracy in reading of the scale of the Mollier diagram.

72  Saturated and Superheated Steam Tables

Figure 5.1  Mollier diagram, SI/Metric units.

Self-Assessment Problems and Questions   73

Chapter 5 Self-Assessment Problems and Questions 1.

Using the saturated liquid enthalpy value for hL and the saturated vapor enthalpy value for hV, at 0.2 MPa and 120.2°C, as listed in the saturated steam tables in Appendix B, calculated the value for hfg.

2.

Calculate the enthalpy of 450 psia and 970°F superheated steam.

3.

Determine the enthalpy of saturated water at 50°C and 1 Bar.

4.

Determine the enthalpy and specific volume for 14 psia steam with a quality of 65%.

6 Phases of Water and Associated Thermodynamics

Topics ■■ ■■

Thermodynamic phases Critical properties of substances, with emphasis on water.

6.1  Phases of Substance Thermodynamic properties and phases of a substance can be defined and determined by two important properties; namely, temperature and pressure. Three most common phases of a substance are as follows: 1.

Solid

2.

Liquid

3.

Gaseous

These three phases, subcategories within these phases, and other pertinent thermodynamic terms are defined below. 6.1.1 Solid The shape and volume of a substance, in solid phase, are non-volatile. A ­substance in its solid phase does not adapt itself to the shape or volume of its container, it retains its original shape and volume. The temperature – under atmospheric pressure conditions - at which a substance attains the solid phase, is called the freezing point. 6.1.2 Liquid The shape of a substance, in liquid phase, is volatile. A substance in its liquid phase adapts itself to the shape of its container. The temperature – under 75

76  Phases of Water and Associated Thermodynamics atmospheric pressure conditions - at which a substance attains the liquid phase is called the melting point. The transition of a substance from solid to liquid phase is, sometimes, referred to as fusion. The term fusion implies the freeing of the atoms or molecules from crystallographic forces and coalescing of the same to form a denser medium, “liquid.” That is the reason for liquid water being denser than ice, and the fact that ice is more buoyant than water. Also, what distinguishes the liquid medium from gaseous medium is that in the liquid medium the surface tension forces tend to retain the molecules of liquid in a contiguous mass or medium. So, while the molecules in liquids – unlike those in solid state (e.g., ice) – may move about, they stay within the bounds of the liquid mass. 6.1.3 Gas The shape as well as volume of a substance, in gaseous phase, are volatile. Gas is a state of matter consisting of a collection of molecular or atomic particles that lacks definite shape or volume. The temperature – under atmospheric pressure conditions - at which a substance attains the gaseous phase, is called the boiling point. When molecules in the liquid phase receive enough energy (thermal), they break free from the forces of surface tension and transcend from the liquid phase to the gaseous phase. 6.1.4  Sensible heat Sensible heat is the heat required, or absorbed, in raising the temperature of a substance, without a change in phase. Example: Heat required to raise the temperature of water from 60°F to 80°F, at sea level, or a pressure of 1 atm. Calculation of sensible heat is demonstrated in other chapters. 6.1.5  Latent heat Latent heat is the heat required or absorbed in changing the phase of a substance. Latent heats for fusion, sublimation and vaporization of water are listed -in SI/Metric Units and US/Imperial Units - in Table 6.1. Consider the following examples as illustration of how to use this table: Example A: Latent Heat of Fusion, hSL, which is the heat required to fully melt ice to liquid water, is 334 kJ/kg. Example B: Latent Heat of Sublimation, hSV, which is the heat required to fully evaporate ice to saturated vapor phase, is 2838 kJ/kg.

6.1  Phases of Substance  77 Table 6.1  Latent heats for water phase transformation, SI units.

  kJ/kg BTU/lbm kcal/kg

Latent Heat Fusion kJ/kg 334 143.4  79.7

Latent Heat Sublimation kJ/kg 2838 1220  677.8

Latent Heat Vaporization kJ/kg 2260 at 100oC  970.3  539.1

Example C: Latent Heat of Vaporization, hfg, which is the heat required to fully evaporate water, at sea level - or a pressure of 1 atm - and 100 oC, is 2260 kJ/kg. 6.1.6  Saturation temperature Saturation temperature, at a given pressure, is the temperature below which a gas or vapor would condense to liquid phase. For example, the saturation temperature at standard atmospheric pressure of 101 kPa (0.1014MPa) is 100°C. See the circled pressure and temperature values in Table 6.2. Now, if the pressure is reduced to 12.4 kPa (0.0124 MPa), the saturation temperature would drop to 50°C. In other words, if the pressure is reduced to 12.4 kPa (0.0124 MPa), the water’s boiling point would be reduced to 50°C. 6.1.7  Saturation pressure Saturation pressure, at a given temperature, is the pressure above which a gas or vapor would condense to liquid phase. For example, the saturation pressure at a temperature of 100°C is 101 kPa (0.1014MPa). See the circled pressure and temperature values in Table 6.1. Now, if the pressure is increased to (1.5547 MPa), the saturation temperature rises to 200°C. In other words, if the pressure is increased to 1,554.7 kPa, the water’s boiling point would double, from 100°C to 200°C. This also means that if the initial temperature and pressure conditions are 100°C is 101 kPa, and the pressure is escalated by almost 15 folds, up to 1,5547 kPa, the water would no longer be in saturated water phase; it will instead fall back into subcooled liquid phase – see the discussion on subcooled liquids in the next section. Same example in the US unit realm would be that of saturation pressure at a given temperature of say 209°F. The saturation pressure at the given temperature of 209°F, as circled in Table 6.2, would be 14 psia. If, however, the pressure is raised, for instance, to 100 psia, the saturation temperature would rise to 328°F. In other words, if the initial conditions are changed such that

Properties of Saturated Steam by Temperature Metric/SI Units Specific Volume m3/kg

Temp. °C

Abs. Press. MPa

20

0.0023

0.001

57.76

50

0.0124

0.001

12.02

100

0.1014

0.001

200

1.5547

0.001

Enthalpy kJ/kg

Sat. Liquid Sat. Vapor Sat. Liquid nL nV hL 83.9

Evap. hfg

Entropy kJ/kg.°K Sat. Vapor Sat. Liquid Sat. Vapor Temp. hV SL SV °C

2454

2537

0.30

8.67

 20

209

2382

2591

0.70

8.07

 50

 1.672

419

2257

2675

1.31

7.35

100

 0.127

852

1940

2792

2.33

6.43

200

78  Phases of Water and Associated Thermodynamics

Table 6.2  Properties of saturated steam by temperature, SI units.

6.1  Phases of Substance  79

while the temperature remains the same, i.e., 209°F, the pressure is increased from 14 psia to 100 psia, the water would fall into the subcooled liquid state. 6.1.8  Subcooled liquid When the temperature of a liquid is less than its boiling point, at a given pressure, it is referred to be in a subcooled state or phase. Example: Water at room temperature (77°F, 25°C), at sea level (1-Atm or 1-Bar), is considered to be subcooled, in that, addition of a small amount of heat will not cause the water to boil.

••

If, at a certain pressure, temperature is a determining variable, a substance is said to be in a subcooled liquid phase when its temperature is below the saturated temperature value corresponding to its pressure.

••

Conversely, if at a certain temperature, pressure is a determining variable, a substance is said to be in a subcooled liquid phase when its pressure is greater than the saturated pressure value corresponding to that temperature.

6.1.9  Saturated liquid When the temperature of a liquid is almost at its boiling point, such that addition of a small amount of heat energy would cause the liquid to boil, it is said to be saturated. In other words, the liquid is saturated with heat and cannot accept additional heat without transcending into vapor phase. Example: Water at 212°F, or 100°C, at sea level. 6.1.10  Saturated vapor Vapor that has cooled off to the extent that it is almost at the boiling point, or saturation point, and at the verge of condensing, is called a saturated vapor. Conversely, one could view saturated vapor point as a point that lies on the saturated vapor line on the temperature enthalpy graph shown in Figure 6.2, where addition of more heat energy to the vapor will drive it into the superheated realm. 6.1.11  Liquid-vapor phase A substance is said to be in a liquid-vapor phase when its temperature is equal to the saturated temperature value corresponding to its pressure. When

80  Phases of Water and Associated Thermodynamics water is in the liquid-vapor phase, in most cases, a portion of the total volume of water has evaporated; the remaining portion is in saturated water state. 6.1.12  Superheated vapor Superheated vapor is vapor that has absorbed heat well beyond its boiling point. Loss of a small amount of heat would not cause superheated vapor to condense.

••

If, at a certain pressure, temperature is the determining variable, then a vapor is said to be in a superheated vapor state when its temperature exceeds the saturation temperature corresponding to its pressure.

••

Conversely, if at a certain temperature, pressure is the determining variable, a vapor is said to be in a superheated vapor state when its pressure is less than the saturation pressure corresponding to that temperature.

Example Problem 1 Answer the following questions for water at a temperature of 153°F and pressure of 4 psia: a) Heat content for saturated water. b) Specific heat (BTU/lbm) required to evaporate the water. c) If the water were evaporated, what would the saturated vapor heat content be? d) What state or phase would the water be in at the stated temperature and pressure? e) What would the entropy of the water be while it is in saturated liquid phase? f) What would the specific volume of the water be while it is in saturated vapor phase? Solution/Answers: a) At 153°F and pressure of 4 psia, the water is in saturated liquid form. According to saturated steam table excerpt in Table 6.3, the saturated water enthalpy is 120.8 BTU/lbm. This value is listed under column

6.1  Phases of Substance  81

labeled hL, in Table 6.3, in the row representing temperature of 153°F and pressure of 4 psia.

Answer: The enthalpy or heat content for saturated water, at the given temperature and pressure, is 120.8 BTU/lbm.

b) The specific heat, in BTU/lbm, required to evaporate the water from saturated liquid phase to saturated vapor phase, is represented by the term hfg. The value of hfg, for saturated water at 153°F and a pressure of 4 psia, as read from Table 6.3 is 1006 BTU/lbm. See circled values in Table 6.3.

Answer: hfg at153°F and 4 psia = 1006 BTU/lbm

c) Saturated vapor heat content, if the water were evaporated, would be the value for hv at153°F and 4 psia, and from Table 6.3 this value is 1127 BTU/ lbm. Answer: hv at153°F and 4 psia = 1127 BTU/lbm d) The water would be in saturated liquid phase at the stated temperature and pressure. All stated saturation temperatures and pressures, in the saturated steam tables, represent the current state of water in saturated liquid phase. Answer: Saturated liquid phase e) The entropy of water at 153°F and a pressure of 4 psia, in saturated liquid phase, as read from Table 6.3, would be sL= 0.22 BTU/(lbm°R). Note that sL value is retrieved form the table and not the sV value. This is because the problem statement specifies the liquid phase. Answer: sL= 0.22 BTU/(lbm.°R) f) The specific volume of water at 153°F and a pressure of 4 psia, in saturated vapor phase, as read from Table 6.3 would be nV = 90 ft3/lbm. Note that nV value is retrieved form the table and not the nL value. This is because the problem statement specifies the vapor phase. Answer: nV = 90 ft3/lbm Same example in the US unit realm would be that of saturation temperature at standard atmospheric pressure of 14 psia (1 Bar). The saturation temperature at standard atmospheric pressure of 14 psia (1 Bar), as circled in Table 6.2, would be 209°F, which is approximately the same as the commonly known boiling point of water at 212°F. Converse to the last example, if the pressure

Properties of Saturated Steam by Pressure US/Imperial Units Abs. Press. psia

Specific Volume ft3/lbm Temp. in °F

Enthalpy Btu/lbm

Sat. Liquid Sat. Vapor Sat. Liquid nL nV hL

Evap. hfg

Entropy Btu/(lbm.°R)

Abs. Sat. Vapor Sat. Liquid Sat. Vapor Press. hV SL SV psia

1036

1105

0.133

1.978

1.0

120.8

1006

1127

0.220

1.862

4.0

38.42

161.2

982

1143

0.2836

1.788

10.0

0.0167

28.0

177.6

972

1149

0.308

1.761

14.0

0.0177

4.4

298.5

889

1188

0.474

1.603

1.0

102

0.0161

333.

4.0

153

0.0164

90.0

10.0

193

0.0166

14.0

209

100

328

69.73

100

82  Phases of Water and Associated Thermodynamics

Table 6.3  Properties of Saturated Steam By Pressure, US units.

6.2  Phase Transformation of Water at Constant Pressure  83

is raised to 100 psia, the saturation temperature would escalate up to 328°F. In other words, as the pressure is increased to 100 psia, the water would boil at a much higher temperature of 328°F. See the circled temperature and pressure values in the US unit version of the saturated steam table excerpt in Table 6.3. Note that increasing the boiling point of the water by raising the pressure on the surface of the water is the same principle that is employed in pressure cookers. By raising the pressure and the temperature, water’s enthalpy or heat content is raised; thus, accelerating the cooking or decomposition of the contents.

6.2  Phase Transformation of Water at Constant Pressure The process of phase transformation of water, under constant pressure, is illustrated in Figure 6.1. In segment A, to the extreme left in Figure 6.1, water is depicted in subcooled liquid phase, at a certain temperature T°C, well below the saturation temperature at the given pressure. As heat is added to the water, as shown in segment B, the temperature of the water rises to T°C + ΔT°C, thus causing the water to expand and achieve the saturated liquid state. This heat, that simply increases the temperature and enthalpy of the water without causing evaporation, is sensible heat. As more heat is added into the system, some of the saturated water transforms into saturated vapor; thus, resulting in a mixture of saturated water and saturated vapor phases, as shown in segment C. The heat added

Figure 6.1  Phase transformation of water at constant pressure.

84  Phases of Water and Associated Thermodynamics

Figure 6.2  Phase transformation of water at constant pressure, depicted in graphical form.

to transform the saturated water into saturated vapor is latent heat. During this phase transformation, the water temperature stays constant, at TSaturation. Further addition of heat leads to the transformation of all of the saturated water into saturated vapor, as shown in segment D. The last stage of this process is represented in segment E of Figure 6.1. This stage shows the transformation of water from saturated vapor state to superheated vapor state, through introduction of more heat. Since the water maintains the vapor state in this stage, the heat added is sensible heat. The phase transformation process described above can also be followed in graphical form as shown in Figure 6.2 below. For instance, the state of water represented by segment A, in Figure 6.1, lies in the subcooled region, to the left of the graph in Figure 6.2. Water represented by segment B lies directly on the saturated water line in Figure 6.2. Water represented by segment C lies within the “bell curve,” shown as the shaded region on the graph in Figure 6.2. Water represented by segment D lies directly on the saturated vapor line. As more heat is added and water transitions into state represented by segment E, it enters the superheated vapor region, shown to the right of the saturated vapor curve on Figure 6.2. Phase and state transformation of water can also be viewed, from physical perspective, as shown in Figure 6.3. This diagram depicts the transformation of water from solid phase to vapor phase as addition of heat drives the phase transformation process, counterclockwise, from ice to vapor state.

6.3  Types of Phase Transformation  85

Figure 6.3  Phase transformation of water.

Once in vapor phase, removal of heat would drive the phase transformation cycle, clockwise, from vapor to ice phase – unless the transformation constitutes sublimation. Phase transformation from ice directly into vapor, as shown in the left half of the diagram in Figure 6.3, represents the sublimation process. Of course, sublimation can be reversed through removal of heat from vapor, resulting in direct phase transformation from vapor to solid state, or ice.

6.3  Types of Phase Transformation Three common phase transformations, and associated latent heat values, have been discussed in earlier sections of this chapter. In addition to providing tabular cross-referencing between various phases of water, Table 6.4 expands on the topic of phase transformation and lists other types of transformation and phases that don’t commonly involve water. Plasma is stated as one of the phases in Table 6.4. Plasma is, essentially, ionized gas. Plasma, or ionized gas, is used in various processes, including, vaporized metal deposition on substances such as semiconductor substrates – for fabrication of integrated circuits or “chips” – and production of metal coated flat glass panels, also known as “performance glazings.” Performance glazings are often used as energy conservation measures in energy efficient commercial construction including ZEB or “Zero Energy Buildings.” When energy or electrical potential is removed is removed from plasma, the plasma phase descends back to

86  Phases of Water and Associated Thermodynamics Table 6.4  Types of phase transformation.

From Solid Liquid

Solid Solid-Solid Transformation Freezing

Gas Plasma

Deposition N/A

To Liquid Gas Melting or fusion Sublimation

Plasma N/A

N/A

N/A

Condensation N/A

Boiling or evaporation N/a Recombination or deionization

Ionization N/A

the regular gas form. When plasma reverts back to the neutral or regular gas phase, the process is referred to as deionization or recombination. 6.3.1  Ideal gas Any gas that behaves in accordance with the ideal gas laws is said to be an ideal gas. Highly superheated vapor behaves like an ideal gas and is treated as such. This is demonstrated in later chapters of this text where ideal gas law based equations are applied in solving various ideal gas related thermodynamic problems. 6.3.2  Real gas Any gas that does not behave according to the ideal gas laws is said to be a real gas. Saturated vapor is considered real gas. 6.3.3  Critical point If temperature and pressure of a liquid are increased, eventually, a state is reached where liquid and vapor phases co-exist and are indistinguishable. This point is referred to as a “critical point.” See point D in Figure 6.4. From Table 6.5, it is ostensible that the temperatures and pressures at which critical point exists - for most substances in the universe - would require laboratory conditions on Earth, far from ambient conditions we, as human beings, are acclimatized to. 6.3.4  Critical properties Properties, such as temperature, specific volume, density, and pressure of a substance, at the critical point, are referred to as critical properties. See Tables 6.5 and 6.6.

6.3  Types of Phase Transformation  87 Table 6.5  Critical properties of select substances.

Substance Argon Ammonia Bromine Cesium Chlorine Fluorine Helium Hydrogen Krypton Neon Nitrogen Oxygen CO2 H2SO4 Xenon Lithium Mercury Sulfur Iron Gold Aluminum Water

Critical Temperature °C °K −122.4 151 132.4 405.5 310.8 584.0 1,664.85 1,938.00 143.8 417.0 −128.85 144 −267.96 5.19 −239.95 33.2 −63.8 209 −228.75 44.4 −146.9 126 −118.6 155 31.04 304.19 654 927 16.6 289.8 2,950 3,220 1,476.9 1,750.1 1,040.85 1,314.00 8,227 8,500 6,977 7,250 7,577 7,850 373.946 647.096

Critical Pressure atm kPa 48.1 4,870 111.3 11,280 102 10,300 94 9,500 76.0 7,700 51.5 5,220 2.24 227 12.8 1,300 54.3 5,500 27.2 2,760 33.5 3,390 49.8 5,050 72.8 7,380 45.4 4,600 57.6 5,840 652 66,100 1,720 174,000 207 21,000 N/A N/A 5,000 510,000 N/A N/A 217.7 22,060

6.3.5  Triple point Triple point of a substance is a state in which solid, liquid and gaseous phases coexist. 6.3.6  Comparison - triple point vs. critical point Both triple point and critical point have been explained earlier sections of this chapter. In this section, we will compare these two points for clarification and distinction. Figure 6.4 is used to illustrate the differences between these two points. The graph depicted in Figure 6.4 shows a plot of pressure as a function of temperature. Note that water – or any other substance – can exist in liquid and gaseous phases, simultaneously, to the right of the line labeled A-B. And, since the critical point “D” lies in the region to the right of line A-B, critical point can only involve two phases, namely, the gaseous phase and the liquid

88  Phases of Water and Associated Thermodynamics Table 6.6  Triple point properties of select substances.

Substance Ammonia Argon Butane Ethane Ethanol Ethylene Hydrogen Hydrogen Chloride Methane Nitric oxide Nitrogen Nitrous Oxide Oxygen Water Zinc

Triple Point Temperature, T (K) 195.4 83.81 134.6 89.89 150 104 13.84 158.96 90.68 109.5 63.18 182.34 54.36 273.16 692.65

Triple Point Pressure, (kPa) 6.076 68.9 7 × 10−4 8 × 10−4 4.3 × 10−7 0.12 7.04 13.9 11.7 21.92 12.6 87.85 0.152 0.6117 0.065

Figure 6.4  Comparison of triple point vs. critical point, a graphical perspective.

Self-Assessment Problems and Questions   89

phase. Also, note that TCr, the temperature at the critical point – or critical temperature - is substantially higher than the, the triple point temperature. At this higher temperature TCr, solid phase cannot exist, for most practical purposes. On the other hand, at the triple point, a substance can exist in all three phases: solid, liquid and gaseous, simultaneously. As shown in Figure 6.4, triple point C lies in the region where solid, liquid, and gaseous phases coexist. The solid phase is to the left of the dashed line E-C, liquid phase in the B-C-D region and vapor phase lies to the right of the line formed by points A, C and D. Another key distinction between triple point and critical point, from pressure point of view, is that critical point pressure, PCr, is significantly higher than the triple point pressure, Pt.

Chapter 6 Self-Assessment Problems and Questions 1.

A boiler is relocated from sea level to a location that is at an elevation of 10,000 ft MSL. Using the table below and Table 6.3, determine the temperature at which the water will boil if the boiler is assumed to be open to atmosphere.

2.

In problem (1), if the objective is just to heat the water close to the boiling point, will the boiler consume more or less fuel than it did when it was located at the sea level?

3.

Answer the following questions for water at a temperature of 193°F and pressure of 10 psia:



a) Heat content for saturated water.



b) Specific heat (BTU/lbm) required to evaporate the water.



c) If the water were evaporated, what would the saturated vapor heat content be?



d) What state or phase would the water be in at the stated temperature and pressure?



e) What would the entropy of the water be while it is in saturated liquid phase?



f) What would the specific volume of the water be while it is in saturated vapor phase?

90  Phases of Water and Associated Thermodynamics

g) What would the phase of the water if the pressure is increased to 20 psia while keeping the temperature constant at 193°F?

Altitude With Mean Sea Level as Ref. Feet 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 6000 7000 8000 9000 10000 15000 20000

Meters 0 152 305 457 610 762 914 1067 1219 1372 1524 1829 2134 2438 2743 3048 4572 6096

Absolute Pressure in Hg Column Inches Hg mm Hg Column Column 29.9 765 29.4 751 28.9 738 28.3 724 27.8 711 27.3 698 26.8 686 26.3 673 25.8 661 25.4 649 24.9 637 24 613 23.1 590 22.2 568 21.4 547 20.6 526 16.9 432 13.8 352

Absolute Atmospheric Pressure psia 14.7 14.4 14.2 13.9 13.7 13.4 13.2 12.9 12.7 12.5 12.2 11.8 11.3 10.9 10.5 10.1  8.29  6.75

kg/cm2 1.03 1.01 0.997 0.979 0.961 0.943 0.926 0.909 0.893 0.876 0.86 0.828 0.797 0.768 0.739 0.711 0.583 0.475

kPa 101 99.5 97.7 96 94.2 92.5 90.8 89.1 87.5 85.9 84.3 81.2 78.2 75.3 72.4 69.7 57.2 46.6

7 Laws of Thermodynamics

Topics ■■ ■■

Laws of thermodynamics, principles, and associated formulas Applications of principles and laws of thermodynamics and associated case study

7.1 Introduction This chapter explores major categories of thermodynamic systems based on their interaction and transactions with the surroundings or environment. The three major categories of thermodynamic systems are introduced and key differences between them are explained. Differentiation between the three types of thermodynamic systems is reinforced through tabular cross-referencing of characteristic properties. Since open thermodynamic systems are somewhat more common than other type of thermodynamic systems, a detailed case study - involving a fossil fuel powered steam power generation system - is undertaken and explained in this chapter. This case study highlights the practical significance and application of steam based thermodynamic systems. The detailed discussion and step by step analysis of each thermodynamic process involved provides the energy engineers an opportunity to understand and hone skills associated with thermodynamic analysis of steam based power harnessing and power generating systems. The discussion in this chapter and the case study builds upon and utilize the thermodynamic concepts, principles, laws and computational methods covered in foregoing chapters. For the reader’s convenience - and to make the analysis versatile - the case study in this chapter is presented in SI/metric units as well as US/imperial units.

91

92  Laws of Thermodynamics

7.2  Major Categories of Thermodynamic Systems Thermodynamic systems can be categorized in myriad ways. However, in this chapter we will focus on categorization of thermodynamic systems based on their interaction with the surroundings or environment. From thermodynamic system and environment interface perspective, thermodynamic systems can be categorized as follows: I. Open Thermodynamic Systems II. Closed Thermodynamic Systems III. Isolated Thermodynamic Systems 7.2.1  Open thermodynamic systems Open thermodynamic systems are systems in which, in addition to the exchange of heat energy with the surroundings, mass or matter are free to cross the system boundary. Also, in open thermodynamic systems, work is performed on or by the system. The type of open thermodynamic systems where entering mass flow rate is the same as the exiting mass flow rate is referred to as a Steady Flow Open System. Examples of Steady Flow Open Systems include pumps, compressors, turbines, and heat exchangers. 7.2.2  Closed thermodynamic systems Closed thermodynamic systems are systems in which no mass crosses the system boundary. Energy, however, can cross through the system boundary in form of heat or work. Examples of closed thermodynamic systems include but are not limited to sealed pneumatic pistons and refrigerants in a refrigeration system. 7.2.3  Isolated thermodynamic systems Isolated thermodynamic systems are systems in which no work is performed by or on the system; no heat is added or extracted from the system and no matter flows in or out of the system. Imagine a rigid sealed steel cylinder containing liquid nitrogen. This steel cylinder is heavily insulated and is placed inside another sealed steel container such that cylinder’s walls do not come in contact with the outside steel container. If vacuum is now created between the outer container and the inner gas cylinder, you would have a thermodynamic system that is “isolated” for most practical purposes. In that, there would be

7.3  Laws of Thermodynamics  93 Table 7.1  Thermodynamic system definition, categorization and cross-referencing.

Interaction of Heat, Mass and Work in Thermodynamic Systems Type of Thermodynamic System Open Closed Isolated Mass Flow Across Boundary Yes No No Work Performed on or By the System Yes Yes No Heat Exchange with the Surroundings Yes Yes No

negligible, in any, heat transfer between the liquid nitrogen and the environment or surroundings outside the outer sealed container. The liquid nitrogen is contained in a sealed steel container, with fixed volume; therefore, no work can be performed by the nitrogen (the system) or the environment (e.g., the air) outside the outer container. In addition, because of the containment or isolation attained through the steel cylinder and the outer sealed container, there would be no transfer of mass or matter. Table 7.1 below facilitates comparison and cross-referencing of the three categories of thermodynamic systems. In addition, this table permits an examination of the three types of thermodynamic systems on the basis of three important thermodynamic process attributes, namely: (1) Mass Flow Across the Boundary, (2) Workflow Across the Boundary, and (3) Heat Exchange with the Surroundings.

7.3  Laws of Thermodynamics Engineering discipline and study of science, in general, utilize principles and laws for developing or deriving equations. These equations are mathematical representation of the engineering or scientific principles and laws. Of course, the key purpose for deriving or developing equations is to be able to define or determine the value of unknown entities or unknown variables. In our continued effort to expand our list of tools for thermodynamic system analyses, at this juncture, we will explore the first law of thermodynamics. 7.3.1  First law of thermodynamics systems The first law of thermodynamics is a statement of the law of conservation of energy in the thermodynamics realm. In other words, the net energy entering a thermodynamic system is equal to the net change in the internal energy of the system plus the work performed by the system. All energy and work are accounted for at all points in the thermodynamic system. Of course, energy status at one point is compared with another point in the system as the laws of

94  Laws of Thermodynamics thermodynamics are applied to derive equations for system or process analysis, in accordance with law of conservation of energy. As we transform the first law of conservation of energy into an equation with practical application and significance, it is important to reiterate that energy, heat and work are mathematically equivalent. Energy, heat and work can, therefore, be added or subtracted linearly in an equation. Mathematical Statement of the First Law of Thermodynamics in a Closed Thermodynamic System is as follows: Q = ΔU + ΔKE + ΔPE + W Where, ΔU ΔKE ΔPE Q W

(7.1)

= Change in Internal Energy = Change in Kinetic Energy = Change in Potential Energy = Heat energy entering (+), or leaving (-) the system = Work performed by the system on the surroundings is positive and work performed by the surroundings on the system is negative.

Mathematical Statement of the First Law of Thermodynamics in an Open Thermodynamic System, also referred to as SFEE, or Steady Flow Energy Equation – like the application of the First Law of Thermodynamics in a Closed Thermodynamic System - would be as follows: Q = ΔU + ΔFE + ΔKE + ΔPE + W Where, ΔU ΔFE ΔKE ΔPE Q W

(7.2)

= Change in Internal Energy = Change in Flow Energy = Δ(PV) = Change in Kinetic Energy = Change in Potential Energy = Heat energy entering (+), or leaving (-) the system = Work performed by the system on the surroundings is positive and work performed by the surroundings on the system is negative.

The Steady Flow Energy Equation, or the SFEE, representing the application of the First Law of Thermodynamics in an Open Thermodynamic System can be written in the specific, or per unit mass, form as: q = Δu + ΔFEspecific + ΔKEspecific + ΔPEspecific + wshaft(7.3)

7.3  Laws of Thermodynamics  95

Figure 7.1  Open thermodynamic system with a turbine.

Since, Δh = Δu + ΔFEspecific(7.4) The SFEE can be written, in a more practical form, for a turbine open system as: q = Δh + ΔKEspecific + ΔPEspecific + wshaft (7.5) Or, q = (h2 – h1) + 1/2 (v22 – v12) + g(z2 – z1) + wshaft (7.6) This Steady Flow Energy Equation is stated in the SI or metric realm. Where, h1 h2 v1 v2 z1 z2 q

wshaft Wshaft g

= Enthalpy of the steam entering the turbine, in kJ/kg. = Enthalpy of the steam exiting the turbine, in kJ/kg. = Velocity of the steam entering the turbine, in m/s. = Velocity of the steam exiting the turbine, in m/s. = Elevation of the steam entering the turbine, in meters. = Elevation of the steam exiting the turbine, in meters. = Specific heat added or removed from the turbine system, in kJ/kg. The value of this variable is “+” when heat is being added to the thermodynamic system and it is “–” when heat is being lost by the thermodynamic system. = Specific work or work per unit mass; measured in kJ/kg. = Absolute work performed by the turbine shaft, measured in kJ. = Acceleration due to gravity, 9.81 m/s2

96  Laws of Thermodynamics Note, Pshaft = wshaftṁ(7.7) Where, ṁ = Mass flow rate of the system, in kg/s or lbm/s. The SI version of the SFEE, thus derived, can be written in a more useful form for power calculation in turbine type open thermodynamic systems as: ṁ(q) = ṁ[(h2 – h1) + 1/2(v22 – v12) + g(z2 – z1)] + ṁwshaft (7.8) Since, Q̇ = ṁ(q) 

(7.9)

Where, Q̇ = Flow rate of heat added or removed from the turbine system, in kJ/second. Therefore, the SI version of the SFEE can be written, in a more practicle form, for a power calculation in turbine type open system as: Q̇ = ṁ [(h2 – h1) + 1/2(v22 – v12) + g(z2 – z1)] + Pshaft 

(7.10)

This Open System Steady Flow Energy Equation can be written, for application in the US or Imperial unit realm, as: ṁ(q ) = ṁ [(h2 – h1) + 1/2(v22 – v12)/(gcJ) + g(z2 – z1)/(gcJ)] + ṁwshaft (7.11) Or, Q̇ = ṁ [(h2 – h1) + 1/2(v22 – v12)/(gcJ) + g(z2 – z1)/(gcJ)] + Pshaft (7.12) Where, h1 h2 v1

= Enthalpy of the steam entering the turbine, in BTU/lbm. = Enthalpy of the steam exiting the turbine, in BTU/lbm. = Velocity of the steam entering the turbine, in ft/s.

7.3  Laws of Thermodynamics  97

v2 z1 z2 q Q̇ wshaft Wshaft ṁ g gc J

= Velocity of the steam exiting the turbine, in ft/s. = Elevation of the steam entering the turbine, in ft. = Elevation of the steam exiting the turbine, in ft. = Specific heat added or removed from the turbine system, in BTU/lbm = Flow rate of heat added or removed from the turbine system, in BTUs/s. = Specific work or work per unit mass; measured in BTU/lbm. = Absolute work performed by the turbine shaft, measured in BTUs. = Mass flow rate of the system, in lbm/s. = Acceleration due to gravity, 32.2 ft/s2 = Gravitational constant, 32.2 lbm-ft/lbf s2 = BTU to ft lb conversion factor = 778 ft lbf/BTU

7.3.2  Second law of thermodynamics The Second Law of Thermodynamics can be stated in multiple ways. Some of the more common and practical statements of the Second Law of Thermodynamics are, briefly, discussed in this section The Second Law of Thermodynamics is also known as the law of increasing entropy. While quantity of total energy remains constant in the universe as stipulated by the First Law of Thermodynamics, the Second Law of Thermodynamics states that the amount of usable, work producing, energy in the universe continues to decline; irretrievably lost in form of unusable energy. Since entropy is defined as unusable energy, the Second Law of Thermodynamics can be interpreted to state that in a closed system, such as the universe, entropy continues to increase. In other words, the second law of thermodynamics states that the net entropy must always increase in practical, irreversible cyclical processes. The second law of thermodynamics can also be stated mathematically in form of Eq. 7.13, below: δQ (7.13) T Equation 7.13 stipulates that the increase in entropy of a thermodynamic system must be greater than or equal to the integral of the incremental heat absorbed, divided by the temperature during each incremental heat absorption. ΔS ≥

∫

98  Laws of Thermodynamics Kevin-Planck statement of the second law of thermodynamics, effectively, implies that it is impossible to build a cyclical engine that has an ­efficiency of 100%.

7.4  Case Study 7.1: SI Units Technical feasibility of a toping cycle cogenerating power plant is being studied at Station “Zebra.” This facility is to be stationed in a remote Arctic region. The objective of this plant is to produce steam and generate electricity for an Arctic Environmental Monitoring and Deep Sea Mining Facility. Due to saltwater corrosion risk, it has been established that local glacial ice will be harvested and utilized for steam production purposes. Natural gas boilers are to be used to heat ice, melt it, and generate steam. The average temperature of the glacier ice hovers at -10°C, through the year. The glacial ice is to be, ultimately, melted and converted to 500°C, 2.5 MPa, steam; the steam enters the turbine at this temperature and pressure. The steam is discharged by the turbines at 150°C and 50 kPa. Although some of the hot water and discharged steam are always needed for the mining process and to heat the station, for simplicity sakes, the quantities are assumed to be negligible. The condensate, under worst case scenario, is also required to serve various utility and process water needs in the station. Assume that the turbine represents an open, steady flow, thermodynamic system. In other words, the SFEE, Steady Flow Energy Equation applies. Station Zebra, in worst case scenario, would operate on a 24/7 schedule at full output of 10 MW. a) Estimate the mass flow rate ṁ for generation of 10 MW electrical power ̇ is 30 kJ/s (or 0.03 MW), exit if the rate of turbine casing heat loss, Q, velocity of steam, v2 , is 35 m/s, entrance velocity of steam, v1 , is 15 m/s, steam exit elevation is z2 = 1m, steam intake elevation is z1 = 0.5m. Assume electric power generator efficiency of 90%. Extrapolate the answer into approximate truck loads per hour. Assume truck capacity of approximately 10 cubic meters. b) Assume that the power station is generating 10 MW of electric power. Calculate the amount of total heat energy needed, in BTU’s/hr, to convert -10°C harvested ice to 500°C, 2.5 MPa steam per hour. c) Calculate the volume, in cu-ft, of natural gas required to power up the station, each day. Assume 98% burner efficiency for all burners being used in all stages of the process. d) If the natural gas transportation cost is $4.85/DT, in addition to the well head or commodity cost stated in Table 7.1, what would be the annual

Table 7.1  Energy content and cost comparison of common fuels.

Energy Content and Cost Comparison Fuel Type or Energy Source

Heat Content

MM BTU/ BTU/ CuPricePe1 DT/MCF Gallon * $/DT * DT/gal BTU/gal Barrel ft.

Diesel/No. 2 Fuel Oil No. 6 Fuel Oil Natural Gas Electricity

0.092 91,600

3.35

0.1 38 140,000

5.6

0.144 143,888

6.8

2,488

1,034

1.034

BTU/kWh

$/kWh

$/MM BTU

$1.42

$15.38

27.29

$2.65

$1 9.1 7

23.429

$1.00

$4.15 $ 30.01

3,41 2

0.1 024

* Note: These costs represent a Ja nuary 3, 2011 snapshot of wholesa le or industrial market costs.

4.15 30.01

7.4  Case Study 7.1: SI Units  99

Propane

US Sept. 2010 Average

Approx. Cost

100  Laws of Thermodynamics fuel cost of operating this station? Note: To put gas pricing in perspective, at the time of publication of this text, Henry Hub Natural Gas Spot Price Graph and Trend is depicted in Figure 7.2 (A); courtesy of EIA, US Energy Information Administration. e) What is the overall energy efficiency of the power station? f) If heat is added to the steam turbines, would the steady flow energy process in the turbine system constitute an adiabatic process or a non-adiabatic process? g) What is the change in entropy, Δs, in the turbine system?

7.5  Case Study Solution Strategy Before embarking on the analyses and solution for this case study, let’s highlight some important facts from the case study statement. 1.

As apparent from the case study statement, the working fluid or system consists of water, in various phases.

2.

Water is introduced into the overall thermodynamic system in form of -10°C ice and is then heated during various sensible and latent stages. It is, eventually, fed into the turbine as 500°C, 2.5 MPa superheated steam.

3.

Unlike a typical Rankine Cycle Heat Engine, the steam exhausted from the turbine is not condensed, pumped and recycled through the boilers to repeat the heat cycle. Instead, fresh ice is harvested, melted and introduced as working fluid, under worst case scenario. Even though, under normal operating circumstances, the facility will not require 10MW of electricity and majority of the condensate would be recycled through the turbine, all analyses are performed assuming all steam must be generated, strictly, through harvested ice and entire 10W of electrical power must be available.

4.

For simplicity, the thermodynamic process flow for this system is laid out in Figure 7.2. Note: In an actual project setting, developing such a process flow diagram would constitute the first order of business as an engineer develops the analytical strategy and embarks on analyzing this process.

5.

Even though Figure 7.2 shows process and potable water being removed from the system at Stages 2 and 3, it is assumed to be negligible. Thus,

Figure 7.2  Thermodynamic process flow diagram, case study 7.1.

7.5  Case Study Solution Strategy  101

102  Laws of Thermodynamics

Figure 7.2(A)  natural gas pricing at the time of publication of this text.

supporting the assumption that mass flow rate of the working fluid stays constant through the system. 6.

Fuel heat content and fuel cost information is available through Table 7.1, supported by Figure 7.2 (A).

7.

Specific heat and latent heat data for water and ice are provided through Tables 7.2 and 7.3, for sensible heat and latent heat calculations, respectively.

8.

All of the data pertinent to the application of SFEE equation, at the turbine, is given, including the velocities, temperatures, pressures and elevations.

7.5  Case Study Solution Strategy  103 Table 7.2  Specific heat, cp, for selected liquids and solids.

Approximate Specific Heat, cp, for Selected Liquids and Solids cp in cal/g K or Molar Cp Substance cp in J/g K BTU/lb F J/mol K Aluminum Bismuth Copper Brass Gold Lead Silver Tungsten Zinc Mercury Alcohol(ethyl) Water Ice (-10 C) Granite Glass

0.9 0.123 0.386 0.38 0.126 0.128 0.233 0.134 0.387 0.14 2.4 4.186 2.05 0.79 0.84

0.215 0.0294 0.0923 0.092 0.0301 0.0305 0.0558 0.0321 0.0925 0.033 0.58 1 0.49 0.19 0.2

24.3 25.7 24.5 ... 25.6 26.4 24.9 24.8 25.2 28.3 111 75.2 36.9 ...  

Table 7.3  Latent heat for phase transformation of water.

 

Latent Heat Fusion hsl

Latent Heat Sublimation hig

Latent Heat Vaporization hfg

kJ/kg BTU/lbm kcal/kg

333.5 143.4  79.7

2838 1220  677.8

2257  970.3  539.1

9.

Final output of the power generating station is given in terms of the 10 MW rating of the generator and its stated efficiency of 90%.

The overall thermodynamic process flow can be tiered into stages that involve either sensible heating or latent heating. All heating stages of this comprehensive process are depicted in Figure 7.2. Each stage of the overall process is labeled with pertinent entry and exit temperature and pressure, as applicable. Each stage is named as either Sensible or Latent Stage. Furthermore, each stage states the phase of water at point of entry and exit. a) Estimate the mass flow rate ṁ for generation of 10 MW electrical power output if the rate of turbine casing heat loss, Q̇ , is 30 kJ/s (or 0.03 MW), exit velocity of steam, v2 , is 35 m/s, entrance velocity of steam, v1 , is 15 m/s, steam exit elevation is z2 = 1m, steam intake elevation is z1 = 0.5m.

104  Laws of Thermodynamics Assume electric power generator efficiency of 90%. Extrapolate the answer into approximate truck loads per hour. Assume truck capacity of approximately 10 cubic meters. Solution This part of the case study can be analyzed and solved by simply focusing on the very last stage and applying the SFEE in form of Eq. 7.10. The computation of mass flow rate does not require assessment of the heat required at the various stages of the overall thermodynamic process, in this case study, because of the following key assumption included in the problem statement: “Also assume the potable and process water demand and flow, along the thermodynamic process stages, to be relatively negligible. “ In other words, the mass flow rate is assumed to be constant throughout the process, and any discharge of water during individual stages of the overall process is negligible. The turbine segment of the overall power generating system is illustrated in Figure 7.3, below. The enthalpy values are obtained from the superheated steam table excerpt in Table 7.4. Apply SFEE, in form of Eq. 7.10. Q̇ = ṁ [(h2 – h1) + 1/2(v22 – v12) + g(z2 – z1)] + Pshaft (7.10)

Figure 7.3  Case study 7.1, mass flow rate analysis.

7.5  Case Study Solution Strategy  105

Rearrangement of Eq. 7.10 yields: ṁ = (Q̇ - Pshaft ) / [(h2 – h1) + 1/2(v22 – v12) + g(z2 – z1)] Given: Q̇ Pshaft h2 h1 v2 v1 z2 z1

= - 0.03 MW = - 0.03x106 W = - 0.03x106 J/s = (10 MW) / hg / ht = (10 MW) / (0.95) / (0.95) = 11.08 x106 W = 11.08 x106 J/s = ho = 2780 kJ/kg = 2780x103 J/kg {See Table 7.4} 3 = hi = 3462 kJ/kg = 3462x10 J/kg {See Table 7.4} = 35 m/s = 15 m/s = 1m = 0.5m

Apply Eq. 7.10, in its rearranged form as follows: ṁ = ( Q̇ - Pshaft ) / [(h2 – h1) + 1/2(v22 – v12) + g(z2 – z1)] ṁ = (-0.03 × 106 J/s - 11.08 × 106 J/s ) / [(2780 × 103 J/kg – 3462 × 103 J/kg) + 1/2 {(35m/s)2 – (15m/s)2} + 9.81 m/s2(1m – 0.5m)] ṁ = (-0.03 × 106 J/s - 11.08 × 106 J/s ) / (-682,000 J/kg + 500 J/kg + 4.9J/kg ) ṁ = 16.08 kg/s, Or, ṁ = {(16.08 kg/907.2kg/ton)/s}(3600 s/h) ṁ = 65 tons/h, or Since the Density of ice = 916.8 kg/m3, The Volumetric flow rate, V̇ , would be: V̇ = (16.08 kg/s)(3600 s/h)/( 916.8 kg/m3) V̇ = 64.02 cu-meters/h At an estimated 10 cubic meters per truck load, this volumetric mass flow rate would amount to: V̇ = (64.02 cu-meters /h)/10 Or, V̇ = 6.4 truckloads per hour

106  Laws of Thermodynamics Table 7.4  Excerpt, superheated steam table, SI units.

Abs. Press. in MPa (Sat. T, °C ) 0.05 (81.33)     0.1 (99.61)     1.0 (179.89)     2.5 (223.99)     3.0 (233.86)     4.0 (250.36)  

n h s   n h s   n h s   n h s   n h s   n h s

Properties of Superheated Steam Metric/SI Units “h” in Temp. in “n” in kJ/kg °C m3/kg 150 300 500 3.889 5.284 7.134 2780.1 3075.5 3488.7 7.9401 8.5373 9.1546       1.9367 2.6398 3.5656 2776.6 3074.5 3488.1 7.6147 8.2171 8.8361         0.2580 0.3541   3051.7 3479.0   7.1247 7.7640         0.0989 0.13998   3008.8 3462.1   6.6438 7.3234         0.0812 0.1162   2994.3 3457.0   6.5412 7.2356         0.0589 0.0864   2961.7 3445.8   6.3638 7.0919

“s” in kJ/kg.°K 650                 0.4245 3810.5 8.1557   0.1623 3799.7 7.7056   0.1405 3797.0 7.6373   0.1049 3790.2 7.4989

  800                 0.4944 4156.2 8.5024   0.1896 4148.9 8.0559   0.1642 4147.0 7.9885   0.1229 4142.5 7.8523

b) Assume that the power station is generating 10 MW of electric power. Calculate the amount of heat needed, in BTU’s/hr, to convert -10°C harvested ice to 500°C, 2.5 MPa steam, per hour. Solution Part (b) of this case study does require accounting for heat added during each of the five (5) stages of the overall process. Therefore, this part is divided into

7.5  Case Study Solution Strategy  107

five sub-parts, each involving either sensible or latent heat calculation, based on the entry and exit temperature and phase status. Table 7.2 lists specific heat for water and ice. These heat values will be used in the sensible heat calculations associated with Part (b). Table 7.3 lists latent heat values for water. These values will be used to compute the latent heats associated with stages that involve phase transformation. (i) Calculate the heat required to heat the ice from -10°C to 0°C: Since there is no change in phase involved, the entire heat absorbed by the ice (working substance) in this stage would be sensible heat. The first stage of the overall power generating system is illustrated in Figure 7.4, below. Given: Ti = -10°C Tf = 0°C cice = 2.05 kJ/kg.°K

{Table 7.3}

Utilizing the given information: ΔT = Tf - Ti \ ΔT = 0 – (-10°C) = +10°C Since ΔT represents the change in temperature and not a specific absolute temperature, \ ΔT = +10°C = +10°K

Figure 7.4  Case study 7.1 Stage 1 sensible heat calculation.

108  Laws of Thermodynamics Mathematical relationship between sensible heat, mass of the working substance, specific heat of the working substance and change in temperature can be stated as: Qs(heat ice) = mcice ΔT(7.14) And, Q̇ s(heat ice) = ṁcice ΔT 

(7.15)

Where, Qs(heat ice) = Sensible heat required to heat the ice over ΔT Q̇ s(heat ice) = Sensible heat flow rate required to heat the ice over ΔT m = Mass of ice being heated cice = Specific heat of ice = 2.05 kJ/kg °K = Change in temperature, in°C or°K ΔT ṁ = Mass flow rate of water/ice = 16.30 kg/s Or,

ṁ = (16.30 kg/s)(3600 s/h) = 58,690 kg/h

Then, by application of Eq. 7.15: = (58,690 kg/hr)(2.05 kJ/kg °K)(10°K) Q̇ s(heat ice)

Or, Q̇ s(heat ice) = 1,203,152 kJ/hr Since there are 1.055 kJ per BTU, = (1,203,152 kJ/hr)/(1.055 kJ/BTU) Q̇ s(heat ice)

Or, Q̇ s(heat ice) = 1,140,428 BTU/hr (ii) Calculate the heat required to melt the ice at 0°C. Since change in phase is involved in this case, the heat absorbed by the ice (working substance) in this stage would be latent heat. The 2nd stage of the overall power generating system is illustrated in Figure 7.5, below.

7.5  Case Study Solution Strategy  109

Figure 7.5  Case study 7.1 stage 2 latent heat calculation.

Mathematical relationship between latent heat, mass of the working substance, and the heat of fusion of ice can be stated as: Ql(latent ice) = hsl (ice)m 

(7.16)

And, Q̇ l(latent ice) = hsl (ice)ṁ(7.17) Where, Ql(latent ice) = Latent heat required to melt a specific mass of ice, isothermally Q̇ l(latent ice) = Latent heat flow rate required to melt a specific mass of ice, isothermally, over a period of time m = Mass of ice being melted ṁ = Mass flow rate of water/ice = 60 tons/hr = (60 tons/hr)(907.2 kg/ton) = 58,690 kg/hr, same as part (a) (i) hsl (ice) = Heat of fusion for Ice = 333.5 kJ/kg {Table 7.3} Then, by application of Eq. 7.17: Q̇ l(latent ice) = hsl (ice)ṁ Q̇ l(latent ice) = (333.5 kJ/kg)(58,690 kg/hr) Q̇ l(latent ice) = 19,573,237kJ/hr Since there are 1.055 kJ per BTU, = (19,573,237kJ/hr)/(1.055 kJ/BTU) Q̇ l(latent ice)

110  Laws of Thermodynamics

Figure 7.6  Case study 7.1 stage 3 sensible heat calculation.

Or, Q̇ l(latent ice) = 18,552,831 BTU/hr Note that the specific heat required to melt ice is called heat of fusion because of the fact that the water molecules come closer together as heat is added in the melting process. The water molecules are held apart at specific distances in the crystallographic structure of solid ice. The heat of fusion allows the molecules to overcome the crystallographic forces and “fuse” to form liquid water. This also explains why the density of water is higher than the density of ice. (iii) Calculate the heat reqd. to heat the water from 0°C to 100°C: The 3rd stage of the overall power generating system is illustrated in Figure 7.6, below. Since no phase change is involved in this stage, the heat absorbed by the water in this stage would be sensible heat. Given: = 0°C Ti = 100°C Tf cp-water = 4.19 kJ/kg.°K

{Table 7.3}

Utilizing the given information: ΔT = Tf - Ti \ΔT = 100°C – 0°C = 100°C Since ΔT represents the change in temperature and not a specific absolute temperature, \ ΔT = 100°K

7.5  Case Study Solution Strategy  111

Mathematical relationship between sensible heat, mass of the working substance, specific heat of water (the working substance), and change in temperature can be stated as: Qs(water) = mcp-water ΔT(7.18) And, Q̇ s(water) = ṁcp-water ΔT(7.19) Where, Qs(water) Q̇ s(water) m cp-water ṁ ΔT

= Sensible heat required to heat the water over ΔT = Sensible heat flow rate required to heat the water over ΔT = Mass of water being heated = Specific heat of water = 4.19 kJ/kg °K = Mass flow rate of water = 58,690 kg/hr, as calculated in part (a) = Change in temperature, in °C or °K

Then, by applying Eq. 7.19: Q̇ s(water) = ṁcp-water ΔT Q̇ s(water) = (58,690 kg/hr)(4.19 kJ/kg °K)(100°K) Q̇ s(water) = 24,662,340 kJ/hr Since there are 1.055 kJ per BTU, = (24,662,340 kJ/hr)/(1.055 kJ/BTU) Q̇ s(water)

Or, Q̇

s(water)

= 23,309,254 BTU/hr

(iv) Calculate the heat required to convert 100°C water to 100°C steam: The 4th stage of the overall power generating system is illustrated in Figure 7.7, below. Since change in phase is involved in this case, the heat absorbed by the water in this stage would be latent heat. Mathematical relationship between latent heat of vaporization for water, hfg(water), mass of the water, and the total heat of vaporization of water, Ql(latent , can be stated as: water) Ql(latent water) = hfg (water)m 

(7.20)

112  Laws of Thermodynamics

Figure 7.7  Case study 7.1 stage 4 latent heat calculation.

And, Q̇ l(latent water) = hfg (water) ṁ(7.21) Where, Q l(latent water) = Latent heat of vaporization of water required to evaporate a specific mass of water, isothermally ̇Q = Latent heat of vaporization flow rate required to evapol(latent water) rate a specific mass of water, isothermally, over a given period of time m= Mass of water being evaporated ṁ= Mass flow rate of water = 58,690 kg/hr , same as part (a) (i) hfg (water) = latent heat of vaporization for water = 2257 kJ/kg  {From the steam tables and Table 7.3} Then, by application of Eq. 7.21: =h ṁ Q̇ l(latent water)

fg (water)

Q̇ l(latent water) = (2257 kJ/kg)(58,690 kg/hr) = 132,464,156 kJ/hr Q̇ l(latent water)

Since there are 1.055 kJ per BTU, Q̇ l(latent water) = (132,464,156 kJ/hr)/(1.055 kJ/BTU) Or, Q̇ l(latent water) = 125,558,442 BTU/hr

7.5  Case Study Solution Strategy  113

Figure 7.8  Case study 7.1 stage 5 sensible heat calculation.

(v)  Calculate the heat reqd. to heat the steam from 100°C, 1-atm (102 KPa, or 1-bar) to 500°C, 2.5 MPa superheated steam: The 5th stage of the overall power generating system is illustrated in Figure 7.8, below. Since this stage involves no phase change, the heat absorbed by the steam is sensible heat. In superheated steam phase, the heat required to raise the temperature and pressure of the steam can be determined using the enthalpy difference between the initial and final conditions. Given: Ti = 100°C Pi = 1 - Atm. Note: At 100°C, the saturation pressure is 1- Atm, 1-Bar, or 102 kPa Tf = 500°C Pf = 2.5 MPa For the initial and final temperature and pressure conditions stated above, the enthalpy values, as read from saturated steam table excerpt in Table 7.4a and the superheated steam table excerpt in Table 7.4, are as follows: hi = 2676 kJ/kg at 100°C, 1-Atm hf = 3462 kJ/kg at 500°C, 2.5 MPa Equations for determining the heat required to boost the steam from 100°C, 1-Atm to 500°C, 2.5 MPa are as follows: ΔQ steam = (hf - hi )m 

(7.22)

Q̇ steam = (hf - hi )ṁ 

(7.23)

114  Laws of Thermodynamics

Table 7.4a  Excerpt, saturated steam table, SI units.

Properties of Saturated Steam by Temperature Metric/SI Units Specific Volume m3/kg

Enthalpy kJ/kg

Entropy kJ/kg.°K

Temp. °C

Abs. Press. MPa

20

0.002339

0.0010018

57.7610

83.920

2454.1

2537.5

0.2965

8.6661

20

50

0.012351

0.0010121

12.0280

209.34

2382.7

2591.3

0.7038

8.0746

50

100

0.101420

0.0010435

1.6719

419.10

2257.0

2675.6

1.3070

7.3541

100

200

1.554700

0.0011565

0.1272

852.39

1940.7

2792.1

2.3308

6.4303

200

Sat. Liquid Sat. Vapor Sat. Liquid nL nV hL

Evap. hfg

Sat. Vapor Sat. Liquid Sat. Vapor Temp. hV SL SV °C

7.5  Case Study Solution Strategy  115

Where, ΔQ steam = Addition of heat required for a specific change in enthalpy Q̇ steam = Rate of addition of heat for a specific change in enthalpy hi = Initial enthalpy hf = Final enthalpy m = Mass of steam being heated ṁ = Mass flow rate of steam as calculated in part (a) of this case study = 58,690 kg/hr {From Part (a)} Then, by applying Eq. 7.23: = (h - h )ṁ Q̇ steam

f

i

Q̇ steam = (3462 kJ/kg - 2676 kJ/kg)(58,690 kg/hr) = 46,130,628 kJ/hr Q̇ steam

Since there are 1.055 kJ per BTU, Q̇ s(water) = (46,130,628 kJ/hr)/(1.055 kJ/BTU) Or, Q̇ s(water) = 43,725,713.57 BTU/hr After assessing the heat added, per hour, during each of the five (5) stages of the steam generation process, add all of the heat addition rates to compile the total heat addition rate for the power generating station. The tallying of total heat is performed in BTU’s/hr as well as kJ/hr. Total Heat Addition Rate in kJ/hr: Total Heat Required to Generate 500°C, 2.5 MPa steam from -10°C Ice, at 58,690 kg/hr = 1,203,153 kJ/hr +19,573,237 kJ/hr +24,591,263 kJ/hr + 132,464,156 kJ/hr + 46,130,628 kJ/hr = 223,962,437 kJ/hr Total Heat Addition Rate in BTU’s/hr: Total Heat Required to Generate 500°C, 2.5 MPa steam from -10°C Ice, at 58,690 kg/hr = 1 ,140,429 BTU/hr + 18,552,831 BTU/hr + 23,309,254 BTU/hr + 125,558,442 BTU/hr + 43,725,714 BTU/hr = 212,286,670 BTU/hr

116  Laws of Thermodynamics c) Calculate the volume, in cu-ft, of natural gas required to power up the station, each day. Assume 98% burner efficiency. Solution This part of Case Study 7.1 requires computation of the amount (volume) of natural gas required to power up the station each day. This calculation is straight forward after the derivation of the total energy required, per hour, in part (b). However, the hourly energy requirement must be “scaled up” to account for the 98% efficiency of the boiler burner. The hourly energy requirement, in kJ or BTU’s, can be extended into daily usage. The daily energy usage can then be converted into the volume of natural gas required, based on natural gas energy content listed in Table 7.1. Total Energy Required Per Day

= (212,286,670 BTU/hr)/(0.98) × 24 hr = 5,198,857,236 BTU’s

Since Natural Gas Energy Content = 1034 BTU/cu ft, Total volume of natural gas required per day = (5,198,857,236 BTU’s) / 1034 BTU/cu ft = 5,027,908 cu ft d) If the natural gas transportation cost is $ 4.85/DT in addition to the well head or commodity cost stated in Table 7.1, what would be the annual fuel cost of operating this station? Solution Part (d) of the case study relates to the computation of total annual cost of fuel for the power generation station. This requires the extrapolation of daily energy consumption into annual energy consumption. The annual energy consumption is then multiplied with the total, delivered, cost rate in $/DT to obtain the annual cost in dollars. From part (b), the heat energy required for operating the power station = 212,286,670 BTU/hr Then, based on 24/7 operating schedule assumption, the total energy required per year would be: = (212,286,670 BTU/hr)/(0.98) × 8760 h/year = 1,897,582,891,137, or, approx. 1.900 E+12 BTU’s

7.5  Case Study Solution Strategy  117

Since there are 1,000,000 BTU’s, or 1 MMBTU, per DT, the Total Annual Energy Consumption in DT would be: = 1.900 E+12 BTU’s / (1,000,000 BTU’s/DT) = 1.9 Million DT’s The total cost of natural gas, per DT = Cost at the Source/Well + Transportation Cost = $4.15/DT + $4.85/DT = $9.00/DT Then, the Total Annual Cost for Producing 10MW of Power with Natural Gas = (1.9 Million DT)($9.00/DT) = $ 17.1 Million e) What is the overall energy efficiency of the power station ? Solution/Answer Part (e) entails determination of the overall efficiency of the power generating station. Efficiency calculation, in this case study, requires knowledge of the total electrical energy (or power) produced and the total energy (or power) consumed through the combustion of natural gas. The output of the overall power generating system is ostensible from the given system output rating of 10 MW. The total heat consumption by the system is derived in parts (b), (c) and (d). The overall efficiency of the system would then be assessed by dividing the output power (or energy) by the input power (or energy). From Part (c): Total Energy Required Per Day, in BTU’s = 5,198,857,236 BTU’s Since there are 1.055kJ per BTU, The Total Energy Input, Each Day, in kJ = 5,198,857,236 BTU’s × 1.055 kJ/BTU = 5,484,794,384 kJ This fuel energy usage can be converted into Watts of Mega Watts (MW) by dividing the energy usage by the total number of seconds in a day. This is because the 5,484,794,384 kJ of energy is used over a period of a day. \The System Power Input = (5,484,794,384 kJ)/(24 hr × 3600 s/hr) = 63,480 kJ/s or 63,480 kW

118  Laws of Thermodynamics Since there are 1000 kW per MW, The System Power Input, in MW = (63,480 kW)/(1,000 kW/MW) = 63.48 MW Since Power Output = 10.00 MW, { Given } Total Station Energy Efficiency, in Percent = (Power Output / Power Input) × 100 = 10 MW/ 63.48 MW × 100 = 16% f) If heat is added to the steam turbines, would the steady flow energy process in the turbine system constitute an adiabatic process or a non-adiabatic process? Solution/Answer The answer to Part (f) of the case study lies simply in the definition for adiabatic process. Adiabatic process is a thermodynamic process in which no heat either enters or leaves the thermodynamic system boundary. In this case study, we assumed that heat is lost through the turbine casing. Therefore, this process is a Non-Adiabatic Process. g) What is the change in entropy, Δs, in the turbine system? Solution/Answer Part (g) involves computation of the change in entropy in the turbine segment of the overall thermodynamic process. As characteristic of the turbine stage of a typical heat engine cycle, there is a small change in entropy as the working fluid travels from the intake to the exit point of the turbine. This is evident as the entropy is assessed, through the superheated steam tables, for the incoming steam at 500°C and 2.5MPa, and the outgoing steam at 150°C and 50 kPa. To determine the change in entropy, Δs, in the turbine system we need to identify s1, the entropy of 500°C, 2.5 MPa steam entering the steam turbine and s2, the entropy of 150°C, 50 kPa steam exiting the turbine. Δs = s1 - s2 s1 = 7.3234 kJ/kg °K {From superheated steam tables, See Table 7.4} s2 = 7.9401 kJ/kg °K {From superheated steam tables, See Table 7.4}

7.6 Case Study 7.1: US/Imperial Units 119

∴ Δs = s1 - s2 = 7.3234 kJ/kg °K - 7.9401 kJ/kg °K = - 0.6167 kJ/kg °K Note: The negative sign signifies the loss of heat through the turbine casing.

7.6 Case Study 7.1: US/Imperial Units, With Illustration of Interpolation Method Technical feasibility of a toping cycle cogenerating power plant is being studied at Station “Zebra.” This facility is to be stationed in a remote Arctic region. The objective of this plant is to produce steam and generate electricity for an Arctic Environmental Monitoring and Deep Sea Mining Facility. Due to saltwater corrosion risk, it has been established that local glacier ice will be harvested and utilized for steam production purposes. A natural gas boiler is to be used to generate steam. The average temperature of the glacier ice hovers at 14°F, through the year. The glacier ice is to be melted and converted to 932°F, 362.6 psia, steam; the steam enters the turbine at this temperature and pressure. The steam is discharged by the turbines at 302°F and 7.252 psia. This discharged steam is used for the mining process and to heat the station. The condensate is used as potable, utility and process water. Assume that the turbine represents an open, steady flow, thermodynamic system. In other words, the SFEE, Steady Flow Energy Equation applies. Also assume the potable water demand and flow, along the thermodynamic process stages to be relatively negligible. Station Zebra would operate on a 24/7 schedule. For simplicity, the thermodynamic process flow for this system is laid out in Figure 7.9. Note: In an actual project setting, developing such a process flow diagram would constitute the first order of business as an energy engineer begins analyzing this process. a) Estimate the mass flow rate ṁ for generation of 10 MW electrical power if the rate of turbine casing heat loss, Q̇  , is 28.435 BTU/s (or 0.03 MW), exit velocity of steam, v2, is 114.83 ft/s, entrance velocity of steam, v1, is 49.21 ft/s, steam exit elevation is z2 = 3.28ft, steam intake elevation is z1 = 1.64 ft. Assume electric power generator efficiency of 90%. Extrapolate the answer into approximate truck loads per hour. Assume truck capacity of approximately 353.15 ft3. b) Assume that the power station is generating 10 MW of electric power. Calculate the amount of total heat energy needed, in BTU’s/hr, to convert 14°F harvested ice to 932°F, 362.6 psia steam per hour.

Figure 7.9  Thermodynamic process flow diagram, Case study 7.1.

120  Laws of Thermodynamics

7.7  Case Study Solution Strategy  121

c) Calculate the volume, in cu ft, of natural gas required to power up the station, each day. Assume 98% burner efficiency. d) If the natural gas transportation cost is $4.85/DT in addition to the well head or commodity cost stated in Table 7.5, what would be the annual fuel cost of operating this station? e) What is the overall energy efficiency of the power station? f) If heat is added to the steam turbines, would the steady flow energy process in the turbine system constitute an adiabatic process or a non-adiabatic process? g) What is the change in entropy, Δs, in the turbine system? Before embarking on the analyses and solution for this case study, let’s highlight some important facts from the case study statement.

7.7  Case Study Solution Strategy 1.

As apparent from the case study statement, the working fluid or system consists of water, in various phases.

2.

Water is introduced into the overall thermodynamic system in form of 14°F ice and is then heated during various sensible and latent stages. It is, finally, introduced into the turbine as 932°F, 362.6 psia superheated steam.

3.

Unlike a typical Rankine Cycle Heat Engine, the steam exhausted from the turbine is not condensed, pumped and recycled through the boilers to repeat the heat cycle. Instead, fresh ice is harvested, melted and introduced as working fluid.

4.

Even though Figure 7.9 shows potable water being removed from the system at Stages 2 and 3, it is assumed to be negligible. Thus, supporting the assumption that mass flow rate of the working fluid stays constant through the system.

5.

Fuel heat content and fuel cost information is available through Table 7.5.

6.

Specific heat and latent heat data for water and ice are provided through Tables 7.6 and 7.7, for sensible heat and latent heat calculations, respectively.

122  Laws of Thermodynamics

Table 7.5  Energy content and cost comparison of common fuels.

Energy Content and Cost Comparison Fuel Type or Energy Source

MM BTU/ BTU/ CuPricePe1 DT/MCF Gallon * $/DT * DT/gal BTU/gal Barrel ft.

Propane Diesel/No. 2 Fuel Oil No. 6 Fuel Oil Natural Gas Electricity

0.092 91,600

3.35

0.1 38 140,000

5.6

0.144 143,888

6.8

2,488

1,034

US Sept. 2010 Average

Approx. Cost

Heat Content

1.034

BTU/kWh

$/kWh

$/MM BTU

$1.42

$15.38

27.29

$2.65

$1 9.1 7

23.429

$1.00

$4.15 $ 30.01

3,41 2

0.1 024

* Note: These costs represent a Ja nuary 3, 2011 snapshot of w holesa le or industrial market costs.

4.15 30.01

7.7  Case Study Solution Strategy  123 Table 7.6  Specific heat, cp, for selected liquids and solids.

Approximate Specific Heat, cp, for Selected Liquids and Solids cp in cal/g K or BTU/lb F Molar Cp J/mol K Substance cp in J/g K Aluminum Bismuth Copper Brass Gold Lead Silver Tungsten Zinc Mercury Alcohol(ethyl) Water Ice (-10 C) Granite Glass

0.9 0.123 0.386 0.38 0.126 0.128 0.233 0.134 0.387 0.14 2.4 4.186 2.05 0.79 0.84

0.215 0.0294 0.0923 0.092 0.0301 0.0305 0.0558 0.0321 0.0925 0.033 0.58 1 0.49 0.19 0.2

24.3 25.7 24.5 ... 25.6 26.4 24.9 24.8 25.2 28.3 111 75.2 36.9 ...  

Table 7.7  Latent heat for phase transformation of water.

 

Latent Heat Fusion hsl

Latent Heat Sublimation hig

Latent Heat Vaporization hfg

kJ/kg BTU/lbm kcal/kg

333.5 143.4  79.7

2838 1220  677.8

2257  970.3  539.1

7.

Most of the data pertinent to the application of SFEE equation, at the turbine, is given, including the velocities, temperatures, pressures and elevations.

8.

Final output of the power generating station is given in terms of the 10 MW rating of the generator and its stated efficiency of 90%.

The overall thermodynamic process flow can be tiered into stages that involve either sensible heating or latent heating. All heating stages of this comprehensive process are depicted in Figure 7.9. Each stage of the overall process

124  Laws of Thermodynamics is labeled with pertinent entry and exit temperature and pressure, as applicable. Each stage is named as either Sensible or Latent Stage. Furthermore, each stage shows the phase of water at point of entry and exit. a) Estimate the mass flow rate ṁ for generation of 10 MW electrical power output if the rate of turbine casing heat loss, Q̇  , is 28.435 BTU/s (or 0.03 MW), exit velocity of steam, v2, is 114.83 ft/s, entrance velocity of steam, v1, is 49.21 ft/s, steam exit elevation is z2 = 3.28ft, steam intake elevation is z1 = 1.64 ft. Assume electric power generator efficiency of 90%. Extrapolate the answer into approximate truck loads per hour. Assume truck capacity of approximately 353.15 ft3. Solution This part of the case study can be analyzed and solved by simply focusing on the very last stage and applying the SFEE in form of Eq. 7.10. The computation of mass flow rate does not require assessment of the heat required at the various stages of the overall thermodynamic process, in this case study, because of the following key assumption included in the problem statement: “Also assume the potable water demand and flow, along the thermodynamic process stages to be relatively negligible. “ In other words, the mass flow rate is assumed to be constant throughout the process, and any discharge of water during individual stages of the overall process is negligible. The turbine segment of the overall power generating system is illustrated in Figure 7.10, below. The enthalpy values are obtained from the superheated steam table excerpt in Table 7.8. 7.7.1 Single and double interpolation of steam table data for enthalpy determination Interpolation is often required when the retrieving data from tables such as the Saturated Steam Tables or the Superheated Steam Tables. Interpolation, specifically in steam tables, is needed when the given pressure, temperature or both don’t coincide with the standard pressure and temperature values on the given tables. This case study, in the US unit realm, offers an opportunity to study the interpolation method. Even though interpolation method is being illustrated on the basis of steam tables here, this technique can be employed for interpolation of other types of tabular data, as well. In this case study, enthalpy values

7.7  Case Study Solution Strategy  125

Figure 7.10  Case study 7.1, Mass flow rate analysis, US units.

need to be retrieved from the Superheated Steam Tables in Appendix B. The pressure and temperature for the initial (turbine entry) enthalpy, hi, are 362.6 psia and 932°F, respectively. The pressure and temperature for the final (turbine exit) enthalpy, hf, are 7.25 psia and 302°F, respectively. However, as you examine the superheated steam tables for these parameters, exact match for this data is not available in the table. See Tables 7.8 and 7.9, below, for excerpts from the superheated steam tables in Appendix B. As far as the initial point is concerned, hi can be obtained through only one interpolation, or single interpolation; the interpolation associated with the temperature of 932°F, by rounding of the pressure to 360 psia. The magnitude of error in the enthalpy value, due to the 2.6 psia difference between 362.6 psia and 360 psia, is insignificant. This single interpolation approach was adopted in the derivation of the value of hi = 1488.76 BTU/lbm. The formula for single interpolation, applied between the stated or available enthalpy values for 900°F and 1000°F, at 360 psia, is as follows: hi at 932°F, 362.6 psia = hi at 932°F, 360 psia = ((hi at 1000°F, 360 psia - hi at 900°F, 360 psia )/(1000°F -900°F))(932-900) + hi at 900°F, 360 psia Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 7.8 below: hi at 932°F, 362.6 psia = hi at 932°F, 360 psia = ((1525 BTU/lbm – 1471.7 BTU/lbm)/(1000°F - 900°F)) (932°F - 900°F) + 1471.7 BTU/lbm = 1488.76 BTU/lbm

126  Laws of Thermodynamics Note: The available enthalpy values are circled in Table 7.8 below. Double interpolation method is employed in deriving hf, the enthalpy value at 7.25 psia and 302°F. As apparent from the superheated steam tables in Appendix B, this value is not readily available and, therefore, double interpolation must be conducted between the enthalpy values given for 5 psi, 300°F, and 10 psi, 350°F, to derive hf at 7.25 psia and 302°F. Where, hf at 7.25 psia and 302°F is the final enthalpy – enthalpy at the turbine exit point – at 7.25 psia and 302°F. The double interpolation approach, as applied here, will entail three steps. First step involves determination of hf, at 7.25 psia and 300°F, the enthalpy value at 7.25 psia and 300°F. The enthalpy values available and used in this first interpolation step are circled in Table 7.9. The following formula sums up the mathematical approach to this first step: hf at 300°F, 7.25 psia = ((hf at 300°F, 5 psia - hi at 300°F, 10 psia )/(10 psia -5 psia)). (10 psia - 7.25 psia) + hf at 300°F, 10 psia

Table 7.8  Superheated steam table excerpt, US/Imperial units.

Properties of Superheated Steam US/Imperial Units Note: n is in ft3/lbm, h is in BTU/lbm and s is in BTU/(lbm-°R) Temp. °F

Abs. Press. psia (Sat. Temp.°F ) 260 (404.45)   360 (434.43)   450 (456.32)   600 (486.25)  

n h s n h s n h s n h s

500 2.062 1262.5

700 2.5818 1370.8

900 3.0683 1475.2

1000 3.3065 1527.8

1.5901 1.446 1250.6 1.5446 1.1232 1238.9 1.5103      

1.6928 1.8429 1365.2 1.6533 1.4584 1360 1.6253      

1.7758 2.2028 1471.7 1.7381 1.7526 1468.6 1.7117 1.3023 1463.2 1.577

1.8132 2.3774 1525 1.7758 1.8942 1522.4 1.7499 1.411 1518 1.7159

7.7  Case Study Solution Strategy  127

Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 7.9 below: hf at 300°F, 7.25 psia = ((1194.8 BTU/lbm – 1193.8 )/(10 psia -5 psia)). (10 psia - 7.25 psia) + 1193.8 BTU/lbm = 1194.35 BTU/lbm Second interpolation step involves determination of hf, at 7.25 psia and 350°F, the enthalpy value at 7.25 psia and 350°F. The enthalpy values available and used in this interpolation step are circled in Table 7.9. The following formula sums up the mathematical approach associated with this interpolation step: hf at 350°F, 7.25 psia = ((hf at 350°F, 5 psia - hi at 350°F, 10 psia )/(10 psia -5 psia)). (10 psia -7.25 psia) + hf at 350°F, 10 psia Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 7.9 below:

Table 7.9  Superheated steam table excerpt, US/Imperial units.

Properties of Superheated Steam US/Imperial Units Note: n is in ft3/lbm, h is in BTU/lbm Temp. °F and s is in BTU/(lbm-°R)

Abs. Press. psia (Sat. T,°F ) 5 (162.18)   10 (193.16)   15 (212.99)   100 (327.82)  

200 n h s n h s n h s n h s

      38.851 1146.4 1.7926            

300 90.248 1194.8 1.937 44.993 1193.8 1.8595 29.906 1192.7 1.8137      

350 96.254 1218 1.9665 48.022 1217.2 1.8893            

500             37.986 1287.3 1.9243 5.5875 1279.3 1.7089

128  Laws of Thermodynamics hf at 350°F, 7.25 psia = ((1218 BTU/lbm – 1217.2 )/(10 psia -5 psia)). (10 psia -7.25 psia) + 1217.2 BTU/lbm = 1217.64 BTU/lbm The final step in the double interpolation process, as applied in this case, involves interpolating between hf at 300°F, 7.25 psia and hf at 350°F, 7.25 psia the enthalpy values derived in the first two steps above, to obtain the desired final enthalpy hf at 302°F, 7.25 psia . The formula for this final step is as follows: hf at 302°F, 7.25 psia = ((hf at 350°F, 7.25 psia - hf at 300°F, 7.25 psia )/( 350°F - 300°F)) ( 302°F - 300°F) + hf at 300°F, 7.25 psia Substituting enthalpy values derived in the first two steps above: hf at 302°F, 7.25 psia = ((1217.64 BTU/lbm – 1194.35 BTU/lbm)/( 350°F - 300°F)). (302°F - 300°F) + 1194.35 BTU/lbm hf at 302°F, 7.25 psia = 1195.28 BTU/lbm With the key enthalpy values for turbine entry and exit points identified, we can proceed with the computation of the mass flow rate. The Open System Steady Flow Energy Equation for power computation in the US or Imperial unit realm is: Q̇ = ṁ[(h2 – h1) + 1/2(v22 – v12)/(gcJ) + g(z2 – z1)/(gc.J) ] + Pshaft 

(7.12)

Rearrangement of Eq. 7.12 yields: ṁ = (Q̇ - Pshaft ) / [(h2 – h1) + 1/2(v22 – v12)/(gcJ) + g(z2 – z1)/(gcJ) ] (7.24) Where, h1 = Enthalpy of the steam entering the turbine, in BTU/lbm. h2 = Enthalpy of the steam exiting the turbine, in BTU/lbm. v1 = Velocity of the steam entering the turbine, in ft/s. v2 = Velocity of the steam exiting the turbine, in ft/s. z1 = Elevation of the steam entering the turbine, in ft. z2 = Elevation of the steam exiting the turbine, in ft.

7.7  Case Study Solution Strategy  129

Q̇ = Flow rate of heat added or removed from the turbine system, in BTUs/s. ṁ = Mass flow rate of the system, in lbm/s. g = Acceleration due to gravity, 32.2 ft/s2 gc = Gravitational constant = 32.2 lbm ft/lbf s2 J = 778 ft-lbf/BTU Given: Q̇ = - 0.03 MW = - 0.03 × 106 W = - 0.03 × 106 J/s Since there are 1055 Joules per BTU, Q̇ = (- 0.03x106 J/s)/ (1055 J/BTU) = - 28.4 BTU/s Pshaft = (10 MW) / hg / ht = (10 MW) / (0.95) /(0.95) = 11.11 × 106 W (approximately) = 11.11 × 106 J/s Since there are 1055 Joules per BTU, Pshaft = (11.11 × 106 J/s )/(1055 J/BTU) = 10,530 BTU/s h2 h1 v2 v1 z2 z1

= 1195 BTU/lbm, from Appendix B and interpolation stated above. = 1489 BTU/lbm from Appendix B and interpolation stated above. = 114.83 ft/s = 49.21 ft/s = 3.28ft = 1.64 ft

By applying Eq. 7.24: ṁ = (Q̇ - Pshaft ) / [(h2 – h1) + 1/2(v22 – v12)/(gc.J) + g(z2 – z1)/(gc.J) ] (7.24) ṁ = (-28.4 BTU/s - 10,530 BTU/s ) / [(1195 BTU/lbm – 1489 BTU/lbm) + 1/2((114.83 ft/s )2 – (49.21 ft/s) 2)/( 32.2 lbm ft/lbf s2 778 ft lbf/BTU) + (32.2 ft/s2)(3.28ft – 1.64 ft)/ (32.2 lbm ft/lbf s2 778 ft lbf/BTU)]

130  Laws of Thermodynamics For clarity, the ṁ computation equation, Eq. 7.24, with known values substituted, can be stated alternatively as: ( -28.4 BTU / s - 10, 530 BTU / s )

.

m=

(3.28 ft - 1.64 ft ) (1144.83 ft / s) - ( 49.21 ft / s)    1195 BTU / lbm - 1489 BTU / lbm ) + 1 / 2 (32.2lbm - ft / lbf - s )(778 ft - lbf / BTU ) + (32.2 ft / s ) (32.2lbm - ft / lbf - s )(778 ft - lbf / BTU )     2

2

2

2

2

ṁ = 35.94 lbm/s, Or, ṁ = (35.94lbm/s)(3600 s/h) ṁ = 129,382 lbm/hr or 58,687 kg/hr Or, ṁ = {(35.94lbm/2000 lbm/ton)/s}(3600 s/h) ṁ = 65 tons/hour The density data for of ice and water is as follows: Density of Water = 1000 kg/m3 or 62.4 lb/cu ft at 4°C. Density of ice = 917 kg/m3 or 57.26 lbm/cu ft Since our objective is to determine the volumetric flow rate of ice, we will introduce the density of ice in determination of the volumetric flow rate. Therefore, the volumetric flow rate, V,̇ would be: V̇ = (35.94 lbm/s)(3600 s/hr)/(57.26 lbm/ft3) V̇ = 2260 cu ft/hr At an estimated 353.15 ft3 per truck load, this volumetric mass flow rate would amount to: V̇ = (2260 cu ft/hr)/ 353.15 ft3 per truck load Or, V̇ = 6.4 truckloads per hour b) Assume that the power station is generating 10 MW of electric power. Calculate the amount of heat needed, in BTU’s/hr, to convert 14°F harvested ice to 932°F, 362.6 psia steam, per hour. Solution Similar to the metric unit version, part (b) of the US unit version of Case Study 7.1 requires accounting for heat added during each of the five (5) stages of the overall process. Therefore, this part is divided into five sub-parts, each

7.7  Case Study Solution Strategy  131

involving either sensible or latent heat calculation, based on the entry and exit temperature and phase status. Table 7.6 lists specific heat for water and ice. These heat values will be used in the sensible heat calculations associated with Part (b). Table 7.7 lists latent heat values for water. These values will be used to compute the latent heats associated with stages that involve phase transformation. (i) Calculate the heat required to raise the temperature of the ice from 14°F to 32°F. Since there is no change in phase involved, the entire heat absorbed by the ice (working substance) in this stage would be sensible heat. First stage of the overall power generating system is illustrated in Figure 7.11, below. Given: Ti = 14°F Tf = 32°F cice = 0.49 BTU/lbm °F

{Table 7.6}

Utilizing the given information: ΔT = Tf - Ti \ ΔT = 32 – 14°F = 18°F Since ΔT represents the change in temperature and not a specific absolute temperature, ΔT = 10°F = 10°R Mathematical relationship between sensible heat, mass of the working substance, specific heat of the working substance and change in temperature can be stated as: Qs(heat ice) = mcice ΔT(7.14)

Figure 7.11  Case study 7.1, US units, stage 1 sensible heat calculation.

132  Laws of Thermodynamics And, Q̇ s(heat ice) = ṁcice ΔT(7.15) Where, Qs(heat ice) = Sensible heat required to heat the ice over ΔT Q̇ s(heat ice) = Sensible heat flow rate required to heat the ice over ΔT m = Mass of ice being heated cice = Specific heat of ice = 0.49 BTU/lbm °F= 0.49 BTU/lbm °R Since Δ°F = Δ°R = Change in temperature, in°F or°R = 18°F= 18°R ΔT ṁ = Mass flow rate of water/ice = 35.94 lbm/s, or, 129,382 lbm/hr as computed in part (a) Then, by applying Eq. 7.15: Q̇ s(heat ice) = (129,382 lbm/hr)(0.49 BTU/lbm °F)(18°F) Or, Q̇ s(heat ice) = 1,141,149 BTU/hr (ii) Calculate the heat required to melt the ice at 32°F. Since change in phase is involved in this case, the heat absorbed by the ice (working substance) in this stage would be latent heat. The 2nd stage of the overall power generating system is illustrated in Figure 7.12, below. Mathematical relationship between latent heat, mass of the working substance, and the heat of fusion of ice can be stated as: Ql(latent ice) = hsl (ice) m 

(7.16)

And, Q̇ l(latent ice) = hsl (ice) ṁ(7.17)

Figure 7.12  Case study 7.1, US units, stage 2 latent heat calculation.

7.7  Case Study Solution Strategy  133

Where, Ql(latent ice) = Latent heat required to melt a specific mass of ice, isothermally Q̇ l(latent ice) = Latent heat flow rate required to melt a specific mass of ice, isothermally, over a period of time m = Mass of ice being melted ṁ = Mass flow rate of water/ice = 129,382 lbm/hr as ­computed in part (a) hsl (ice) = Heat of fusion for Ice = 143.4 BTU/lbm {Table 7.7} Then, by application of Eq. 7.17: =h ṁ Q̇ l(latent ice)

sl (ice)

Q̇ l(latent ice) = (143.4 BTU/lbm)(129,382 lbm/hr) = 18,553,380 BTU/hr Q̇ l(latent ice)

Note that the specific heat required to melt ice is called heat of fusion because of the fact that the water molecules come closer together as heat is added in the melting process. The water molecules are held apart at specific distances in the crystallographic structure of solid ice. The heat of fusion allows the molecules to overcome the crystallographic forces and “fuse” to form liquid water. This also explains why the density of water is higher than the density of ice. (iii) Calculate the heat reqd. to heat the water from 32°F to 212°F: The 3rd stage of the overall power generating system is illustrated in Figure 7.13, below. Since no phase change is involved in this stage, the heat absorbed by the water in this stage would be sensible heat. Given: Ti = 32°F Tf = 212°F cice = Specific heat of ice = 1.0 BTU/lbm °F = 1.0 BTU/lbm °R {Since Δ°F = Δ°R}

{Table 7.6}

Figure 7.13  Case study 7.1, US units, stage 3 sensible heat calculation.

134  Laws of Thermodynamics Utilizing the given information: ΔT = Tf - Ti \ ΔT = 212°F – 32°F = 180°F Since ΔT represents the change in temperature and not a specific absolute temperature, \ ΔT = 180°R = 180°F Mathematical relationship between sensible heat, mass of the working substance, specific heat of water (the working substance), and change in temperature can be stated as: Qs(water) = mcp-water ΔT(7.18) And, Q̇ s(water) = ṁcp-water ΔT(7.19) Where, Qs(water) Q̇ s(water) m cp-water ṁ ΔT

= Sensible heat required to heat the water over ΔT = Sensible heat flow rate required to heat the water over ΔT = Mass of water being heated = Specific heat of water = 1 BTU/lbm °R = 1 BTU/lbm °F {Table 7.6} = Mass flow rate of water = 129,382 lbm/hr as computed in part (a) = Change in temperature, in °F or °R = 180°F

Then, by applying Eq. 7.19: = ṁc ΔT Q̇ s(water)

p-water

Q̇ s(water) = (129,382 lbm/hr)(1 BTU/lbm °F)(180°F ) = 23,288,760 BTUs/hr Q̇ s(water)

(iv) Calculate the heat required to convert 212°F water to 212°F steam: The 4th stage of the overall power generating system is illustrated in Figure 7.14, below. Since change in phase is involved in this case, the heat absorbed by the water in this stage would be latent heat.

7.7  Case Study Solution Strategy  135

Figure 7.14  Case study 7.1 stage 4 latent heat calculation.

Mathematical relationship between latent heat of vaporization for water, hfg(water), mass of the water, and the total heat of vaporization of water, Ql(latent , can be stated as: water) Ql(latent water) = hfg (water) m 

(7.20)

And, Q̇ l(latent water) = hfg (water) ṁ(7.21) Where, Q l(latent water) = Latent heat of vaporization of water required to evaporate a specific mass of water, isothermally ̇Q = Latent heat of vaporization flow rate required to evapol(latent water) rate a specific mass of water, isothermally, over a given period of time m = Mass of water being evaporated ṁ = Mass flow rate of water = 129,382 lbm/hr, same as part (a) (i) hfg (water) = latent heat of vaporization for water = 970.3 BTU/lbm {From the steam tables and Table 7.7} Then, by application of Eq. 7.21: Q̇ l(latent water) = hfg (water) ṁ Q̇ l(latent water) = (970.3 BTU/lbm)(129,382 lbm/hr) Q̇ l(latent water) = 125,539,355 BTUs/hr (v)  Calculate the heat reqd. to heat the steam from 212°F, 1-atm (102 KPa, or 1-bar) to 932°F, 362.6 psia superheated steam The 5th stage of the overall power generating system is illustrated in Figure 7.15, below. Since this stage involves no phase change, the heat absorbed by the steam is sensible heat.

136  Laws of Thermodynamics

Figure 7.15  Case study 7.1 stage 5 sensible heat calculation.

In superheated steam phase, the heat required to raise the temperature and pressure of the steam can be determined using the enthalpy difference between the initial and final conditions. Given: Ti = 212°F Pi = 1 Atm. Note: At 212°F, the saturation pressure is 1 Atm, 1 Bar, or 102 kPa Tf = 932°F Pf = 362.6 psia For the initial and final temperature and pressure conditions stated above, the enthalpy values - as read from superheated steam table excerpt in Table 7.8, as interpolated in part (a) of this case study, and from Table 7.10 - are as follows: hi at 212°F, 1-Atm = 1149.4 BTU/lbm hf at 932°F, 362.6 psia = hf at 932°F, 360 psia = 1488.76 BTU/lbm Equations for determining the heat required to boost the steam from 212°F, 1-Atm to 932°F , 362.6 psia are as follows: ΔQ steam = (hf - hi )m  = (h - h )ṁ  Q̇ steam

f

i

(7.22) (7.23)

Where, ΔQ steam = Addition of heat required for a specific change in enthalpy Q̇ steam = Rate of addition of heat for a specific change in enthalpy hi = Initial enthalpy hf = Final enthalpy m = Mass of steam being heated ṁ = 129,382 lbm/hr, same as part (a)

Table 7.10  Saturated steam table excerpt, US/Imperial units.

Properties of Saturated Steam by Temperature US/Imperial Units Specific Volume ft3/lbm Temp. °F

1.0

101.69

0.016137

4.0

152.91

0.016356

90.628

14.0

209.52

0.016697

28.048

100

327.82

0.017736

Enthalpy Btu/lbm

Sat. Liquid Sat. Vapor Sat. Liquid nL nV hL 333.51

4.4324

69.728

Evap. hfg

Entropy btu/(lbm.°R)

Abs. Sat. Vapor Sat. Liquid Sat. Vapor Press. hV SL SV psia

1036

1105.4

0.1326

1.9776

1.0

120.89

1006.4

1126.9

0.2198

1.8621

4.0

177.68

972.0

1149.4

0.3084

1.7605

14.0

298.57

889.2

1187.5

0.4744

1.6032

100

7.7  Case Study Solution Strategy  137

Abs. Press. psia

138  Laws of Thermodynamics Then, by applying Eq. 7.23: = (h - h )ṁ Q̇ steam

f

i

Q̇ steam = (1488.76 BTU/lbm - 1149.4 BTU/lbm)(129,382 lbm/hr) = 43,907,076 BTU/hr Q̇ steam

After assessing the heat added, per hour, during each of the five (5) stages of the steam generation process, add all of the heat addition rates to compile the total heat addition rate for the power generating station. The tallying of total heat is performed in BTU’s/hr as well as kJ/hr. Total Heat Addition Rate in BTU’s/hr: Total Heat Required to Generate 932°F, 362.6 psia steam from 1°F Ice, at 129,382 lbm/hr = 1 ,141,149 BTU/hr + 18,553,380 BTU/hr + 23,288,760 BTUs/hr + 125,539,355 BTUs/hr + 43,907,076 BTU/hr = 212,429,719 BTU/hr c) Calculate the volume, in cu-ft, of natural gas required to power up the station, each day. Assume 98% burner efficiency. Solution: This part of Case Study 7.1 requires computation of the amount (volume) of natural gas required to power up the station each day. This calculation is straight forward after the derivation of the total energy required, per hour, in part (b). However, the hourly energy requirement must be “scaled up” to account for the 98% efficiency of the boiler burner. The hourly energy requirement, in kJ or BTU’s, can be extended into daily usage. The daily energy usage can then be converted into the volume of natural gas required, based on natural gas energy content listed in Table 7.5. Total Energy Required Per Day = (212,429,719 BTU/hr)/(0.98) × 24 hr = 5,098,313,255 BTUs Since Natural Gas Energy Content = 1034 BTU/cu-ft, Total volume of natural gas required per day = (5,098,313,255 BTUs) / 1034 BTU/cu-ft = 4,930,670 cu-ft

7.7  Case Study Solution Strategy  139

d) If the natural gas transportation cost is $ 4.85/DT in addition to the well head or commodity cost stated in Table 7.5, what would be the annual fuel cost of operating this station? Solution Part (d) of the case study relates to the computation of total annual cost of fuel for the power generation station. This requires the extrapolation of daily energy consumption into annual energy consumption. The annual energy consumption is then multiplied with the total, delivered, cost rate in $/DT to obtain the annual cost in dollars. From part (b), the heat energy required for operating the power station = 196,909,695 BTU/hr Then, based on 24/7 operating schedule assumption, the total energy required per year would be:

= (212,429,719 BTU/hr)/(0.98) × 8760 h/year = 1.8989 E + 12 BTU’s

Since there are 1,000,000 BTU’s per DT, the Total Annual Energy Consumption in DT would be:

= 1.8989 E+12 BTU’s / (1,000,000 BTU’s/DT) = 1,898,862 DT’s

The total cost of natural gas, per DT = Cost at the Source/Well + Transportation Cost = $4.15/DT + $4.85/DT = $9.00/DT Then, the Total Annual Cost for Producing 10MW of Power with Natural Gas = $ 17,089,754 e) What is the overall energy efficiency of the power station ? Solution/Answer Part (e) entails determination of the overall efficiency of the power generating station. Efficiency calculation, in this case study, requires knowledge of the total electrical energy (or power) produced and the total energy (or power) consumed through the combustion of natural gas. The output of the

140  Laws of Thermodynamics overall power generating system is ostensible from the given system output rating of 10 MW. The total heat consumption by the system is derived in parts (b), (c) and (d). The overall efficiency of the system would then be assessed by dividing the output power (or energy) by the input power (or energy). From Part (c): Total Energy Required Per Day, in BTU’s = 5,098,313,255 BTUs Since there are 1.055kJ per BTU, The Total Energy Input, Each Day, in kJ = 5,098,313,255 BTUs × 1.055 kJ/BTU = 5,378,720,484 kJ This fuel energy usage can be converted into Watts of Mega Watts (MW) by dividing the energy usage by the total number of seconds in a day. This is because the 5,378,720,484 kJ of energy is used over a period of a day. \ The System Power Input = (5,378,720,484 kJ)/(24 hr × 3600 s/hr) = 62,254 kJ/s or 62,254 kW Since there are 1000 kW per MW, The System Power Input, in MW = (62,254 kW )/(1,000 kW/MW) = 62.254 MW Since Power Output = 10.00 MW 

{ Given }

Total Station Energy Efficiency, in Percent = Power Output / Power Input × 100 = 10 MW/ 62.254MW × 100 = 16% f) If heat is added to the steam turbines, would the steady flow energy process in the turbine system constitute an adiabatic process or a non-adiabatic process? Solution/Answer The answer to Part (f) of the case study lies simply in the definition for adiabatic process. Adiabatic process is a thermodynamic process in which no heat either enters or leaves the thermodynamic system boundary. In this case study, we assumed that heat is lost through the turbine casing. Therefore, this process is a Non-Adiabatic Process. g) What is the change in entropy, Δs, in the turbine system?

7.7  Case Study Solution Strategy  141

Solution/Answer Part (g) involves computation of the change in entropy in the turbine segment of the overall thermodynamic process. As characteristic of the turbine stage of a typical heat engine cycle, there is a small change in entropy as the working fluid travels from the intake to the exit point of the turbine. This is evident as the entropy is assessed, through the superheated steam tables, for the incoming steam at 932°F and 362.6 psia, and the outgoing steam at 302°Fand 7.252 psia. To determine the change in entropy, Δs, in the turbine system we need to identify s1, the entropy of 932°F, 362.6 psia steam entering the steam turbine and s2, the entropy of 302°F, 7.252 psia steam exiting the turbine. As evident from superheated steam table excerpts in Tables 7.8 and 7.9, the desired entropy values are not readily available. Interpolation is required to obtain the entropy values at the entry and exit points of the turbine. The formula for single interpolation, applied between the stated or available entropy values for 900°F and 1000°F, at 360 psia, is as follows: si at 932°F, 362.6 psia = si at 932°F, 360 psia = ((si at 1000°F, 360 psia - si at 900°F, 360 psia )/(1000°F   + si at 900°F, 360 psia

- 900°F))(932 - 900)

Substituting entropy values and other given data from superheated steam table excerpt, shown in Table 7.11below: si at 932°F, 362.6 psia = si at 932°F, 360 psia = ((1.7758 BTU/lbm °F – 1.7381 BTU/lbm °F)/(1000°F - 900°F)) (932°F - 900°F) + 1.7381 BTU/lbm °F \ si at 932°F, 362.6 psia = 1.7502 BTU/lbm °F Note: The available entropy values are circled in Table 7.11 below. Double interpolation method is employed in deriving sf, the enthalpy value at 7.25 psia and 302°F. As apparent from the superheated steam tables in Appendix B, this value is not readily available and, therefore, double interpolation must be conducted between the enthalpy values given for 5 psi, 300°F, and 10 psi, 350°F, to derive sf at 7.25 psia and 302°F. Where, sf at 7.25 psia and 302°F is the final enthalpy – enthalpy at the turbine exit point – at 7.25 psia and 302°F. The double interpolation approach, as applied here, will entail three steps. First step involves determination of sf, at 7.25 psia and 300°F, the enthalpy value at 7.25 psia and 300°F. The enthalpy values available and used in this first

142  Laws of Thermodynamics Table 7.11  Superheated steam table excerpt, Us/Imperial units.

Properties of Superheated Steam US/Imperial Units Note: n is in ft3/lbm, h is in BTU/lbm and s is in BTU/(lbm-°R) Temp. °F

Abs. Press. psia (Sat. T,°F ) 260 (404.45)   360 (434.43)   450 (456.32)   600 (486.25)  

n h s n h s n h s n h s

500

700

900

2.062 1262.5

2.5818 1370.8

3.0683 1475.2

1000 3.3065 1527.8

1.5901 1.446 1250.6 1.5446 1.1232 1238.9 1.5103      

1.6928 1.8429 1365.2 1.6533 1.4584 1360 1.6253      

1.7758 2.2028 1471.7 1.7381 1.7526 1468.6 1.7117 1.3023 1463.2 1.577

1.8132 2.3774 1525 1.7758 1.8942 1522.4 1.7499 1.411 1518 1.7159

interpolation step are circled in Table 7.9. The following formula sums up the mathematical approach to this first step: sf at 300°F, 7.25 psia = ((sf at 300°F, 5 psia - si at 300°F, 10 psia )/(10 psia -5 psia)).(10 psia -7.25 psia) + sf at 300°F, 10 psia Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 7.9 below: sf at 300°F, 7.25 psia = ((1.937 BTU/lbm °F – 1.8595 BTU/lbm °F )/(10 psia -5 psia)) (10 psia - 7.25 psia) + 1.8595 BTU/lbm °F = 1.902125 BTU/lbm °F Second interpolation step involves determination of sf, at 7.25 psia and 350°F, the enthalpy value at 7.25 psia and 350°F. The enthalpy values available and used in this interpolation step are circled in Table 7.12. The following formula sums up the mathematical approach associated with this interpolation step: sf at 350°F, 7.25 psia = ((sf at 350°F, 5 psia - si at 350°F, 10 psia )/(10 psia - 5 psia)) (10 psia -7.25 psia) + sf at 350°F, 10 psia

7.7  Case Study Solution Strategy  143 Table 7.12  Superheated steam table excerpt, US/Imperial units.

Abs. Press. psia (Sat. T,°F ) 5 (162.18)   10 (193.16)   15 (212.99)   100 (327.82)  

Properties of Superheated Steam US/Imperial Units Note: n is in ft3/lbm, h is in BTU/lbm Temp. °F and s is in BTU/(lbm-°R) 200 n h s n h s n h s n h s

      38.851 1146.4 1.7926            

300 90.248 1194.8 1.937 44.993 1193.8 1.8595 29.906 1192.7 1.8137      

350 96.254 1218 1.9665 48.022 1217.2 1.8893            

500             37.986 1287.3 1.9243 5.5875 1279.3 1.7089

Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table 7.12 below: sf at 350°F, 7.25 psia = ((1.9665 BTU/lbm °F – 1.8893 BTU/lbm °F )/(10 psia - 5 psia)) (10 psia - 7.25 psia) + 1.8893 BTU/lbm °F = 1.93176 BTU/lbm °F The final step in the double interpolation process, as applied in this case, involves interpolating between sf at 300°F, 7.25 psia and sf at 350°F, 7.25 psia the enthalpy values derived in the first two steps above, to obtain the desired final enthalpy sf at 302°F, 7.25 psia . The formula for this final step is as follows: sf at 302°F, 7.25 psia = ((sf at 350°F, 7.25 psia - sf at 300°F, 7.25 psia )/( 350°F - 300°F)) (302°F - 300°F) + sf at 300°F, 7.25 psia

144  Laws of Thermodynamics Substituting enthalpy values derived in the first two steps above: sf at 302°F, 7.25 psia = ((1.932 BTU/lbm °F – 1.9021 BTU/lbm °F)/( 350°F - 300°F)) ( 302°F - 300°F) + 1.902125 BTU/lbm °F sf at 302°F, 7.25 psia = 1.9033BTU/lbm °F With the key enthalpy values for turbine entry and exit points identified, we can proceed with the computation of Δs: sf at 302°F, 7.25 psia = 1.9033BTU/lbm °F si at 932°F, 362.6 psia = 1.7502 BTU/lbm °F Then, Δs = sf at 302°F, 7.25 psia - si at 932°F, 362.6 psia = 1.7502 BTU/lbm °F - 1.9033BTU/lbm °F = - 0.1531 BTU/lbm °F Note: The negative sign signifies the loss of heat through the turbine casing.

Chapter 7 Self-Assessment Problems and Questions 1.

Why is the efficiency of this power plant in Case Study 7.1 rather low (16%)?

2.

Using the steam tables in Appendix B and the Double Interpolation Method described in Case Study 7.1, US Unit Version, determine the exact enthalpy of a superheated steam at a pressure of 400 psia and temperature of 950°F.

3.

In Case Study 7.1, as an energy engineer you have been retained by Station Zebra to explore or develop an alternative integrated steam turbine and electric power generating system that is capable of generating 10 MW of power with only 60 truckloads, or 54,432 kg, of ice per hour. With all other parameters the same as in the original Case Study 7.1 scenario, determine the total heat flow rate needed, in kJ/hr, to produce 10 MW of electrical power.

4.

If all the working fluid, or steam, discharged from the turbine in Case Study 7.1 is reclaimed, reheated and returned to the turbine, what would be the overall system efficiency?

8 Thermodynamic Processes

Topics ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

Thermodynamic processes Heat engine cycles Steam turbines Temperature-enthalpy diagrams Pressure-enthalpy diagrams Pressure-volume diagrams Temperature-entropy diagrams Practical examples and associated case study.

8.1 Introduction This chapter explores some of the mainstream thermodynamic processes, heat engines and heat engine cycles. Fundamentals of thermodynamic processes, heat engines, heat engine cycles and associated systems are explained and illustrated through process flow diagrams, graphs, tables and pictures. Practical significance, application, analytical methods, and computational techniques associated with heat engine cycles and thermodynamic processes are demonstrated through case study, examples and self-assessment problems.

8.2  Thermodynamic Processes Thermodynamic processes are processes that entail heat, internal energy, enthalpy, entropy, work, pressure, temperature and volume. In this section, we will explore the following thermodynamic processes and illustrate these processes with practical examples: 1.

Adiabatic Process

2.

Isenthalpic Process

145

146  Thermodynamic Processes 3.

Isochoric Process

4.

Isothermal Process

5.

Isobaric Process

6.

Isentropic Process

8.2.1  Adiabatic process Adiabatic process is a thermodynamic process in which no heat either enters or leaves the thermodynamic system boundary. An adiabatic process can also be explained through the following mathematical statements or equations: ΔU = - W 

(8.1)

ΔQ = 0 

(8.2)

Equations 8.1 and 8.2 essentially state that in an adiabatic process, wherein no heat is gained or lost, any work performed on the system or by the system is transformed into a net change in the internal energy of the system. As specifically stated above, Eq. 8.1 represents a scenario where negative work is involved. In other words, work is being performed by the surroundings onto the system. And, since no heat is transferred to or from the environment in an adiabatic process, the work performed by the surroundings onto the system, in this case, is converted into an equivalent amount of increase in the internal energy of the system. This explanation of the adiabatic process is validated by the fact the units for work “W” and internal energy “U” are the same; Joules in the SI/Metric realm and BTU’s in the US/Imperial realm. Equation 8.1 can be restated as: - ΔU = W

(8.1a)

It is important to note that while this restatement of the Eq. 8.1 keeps the equation mathematically equivalent to the original version, the physical significance changes. Equation 8.1a represents a scenario where work is positive and is performed by the system onto the environment or surroundings. Since this is an adiabatic process, there is no transfer or exchange of heat. Therefore, in this case, work is performed by the system, onto the surroundings, at the expense of the internal energy of the system. The negative ΔU signifies a reduction in the internal energy of the system.

8.2  Thermodynamic Processes  147

Work performed in adiabatic processes, such as that performed by the compressor on the refrigerant, would be represented by Eq. 8.3. V2

W=

∫ p.dV (8.3) V1

In a reversible adiabatic process, such as the compression stroke in an internal combustion gasoline engine, the product of pressure and volume is represented as shown in Eq. 8.4. pV  Constant (8.4)

Where, γ = Degrees of freedom of the gas molecules; i.e., 7/5 for nitrogen and oxygen. p = Pressure in SI or US (Imperial) units V = Volume in SI or US (Imperial) units Figure 8.1 represents a “real gas” scenario, where temperature does change to some degree. At an inversion point, however, the temperature does not

Figure 8.1  Work performed in a real gas, adiabatic, thermodynamic process.

148  Thermodynamic Processes

Figure 8.2  Work performed in an ideal adiabatic thermodynamic process.

change during a throttling process. A real gas tends to behave like an ideal gas as temperature approaches the inversion point. Adiabatic process in an “ideal gas” scenario is depicted in Figure 8.2.

8.3 Adiabatic Process Example I – Throttling Process in a Refrigeration System Throttling process in a refrigeration system is an example of an adiabatic process that occurs in the expansion valve; where a high-pressure liquid system (refrigerant) is allowed to expand to a low-pressure liquid, without absorption or release of heat energy.

8.4 Adiabatic Process Example II – Compressor Segment of a Refrigeration System The compression segment of the refrigeration cycle is an adiabatic as well as an isentropic process. During compression the refrigerant, or the system, is compressed from low pressure vapor phase to high pressure vapor phase. No heat is exchanged with the environment during this compression process. The work performed on the refrigerant is negative work, or, “– W.” Since no heat is released by the system, this negative work, in accordance with

8.4  Adiabatic Process Example II – Compressor Segment of a Refrigeration System  149

Eq. 8.1 and the law of conservation of energy, is transformed into internal energy of the refrigerant. Later, when examining pressure-enthalpy graph in figure 8.3, pertaining to a typical refrigerant, you see that the increase in the internal energy of the refrigerant is solely due to the work performed on the refrigerant by the compressor and is evident in for of finite increase in the enthalpy during compression. Compression of vapor, therefore, is an adiabatic process. As discussed later in this text, the compression segment of the refrigeration cycle is not just adiabatic but also an isentropic process. 8.4.1  Isenthalpic or isoenthalpic process An isenthalpic, or isoenthalpic, process is a thermodynamic process in which no change in enthalpy occurs, or Δh = 0 or h1 = h2. A steady-state, steady-flow process, would be isenthalpic if the following conditions are met: 1.

The thermodynamic process is adiabatic – meaning, no heat is exchanged with the environment.

2.

Work is neither performed by the system onto the surroundings nor is it performed by the surroundings onto the system.

3.

There is no change in the kinetic energy of the system or fluid.

8.4.2 Isenthalpic process example – Throttling process in a refrigeration system Refrigeration system throttling process is an example of an isenthalpic process. See Figure 8.3. Throttling of a high-pressure liquid refrigerant to a low pressure liquid phase is an adiabatic process; i.e., no heat is exchanged with the environment. Moreover, no work is done on or by the surroundings, and there is no change in the kinetic energy of the fluid. Note that during the throttling process shown in Figure 8.3, the process adopts a vertical downward path, dropping the pressure precipitously while the enthalpy stays unchanged, thus, rendering the process isenthalpic. In other words, all three requirements or conditions, stated above, for an isenthalpic process are met during the throttling segment of the refrigeration cycle. Other examples of practical isoenthalpic processes include lifting of a relief valve or safety valve on a pressurized vessel. The specific enthalpy of the fluid inside the pressurized vessel is the same as the specific enthalpy of the same fluid immediately after it escapes the vessel. In such a scenario,

150  Thermodynamic Processes

Figure 8.3

Thermodynamic processes in a typical refrigeration cycle.

the temperature and velocity of the escaping fluid can be calculated if the enthalpy is known. In Figures 8.1 and 8.2, an isenthalpic process follows the isotherm line at a specific temperature, and along the isotherm the following relationship between enthalpy, temperature and specific heat holds true: dh = cpdT = 0 Additional examples of isenthalpic process are referenced later in this chapter, under the heat engine cycle discussion. 8.4.3  Constant pressure or isobaric process An isobaric process is a thermodynamic process in which the pressure remains constant. See Figure 8.4, where the curve represents an isobar. Even though the temperature varies as a function of the entropy in this graph, the pressure stays constant.

8.4  Adiabatic Process Example II – Compressor Segment of a Refrigeration System  151

Figure 8.4

Heat absorbed in an isobaric thermodynamic process.

8.4.4 Isobaric process example I: Evaporation stage of a refrigeration cycle Evaporation stage of a refrigeration cycle represents an isobaric process in that the pressure remains constant as the low-pressure liquid system evaporates or changes phases from liquid to gaseous by absorbing the heat energy from the “return” air passing through the heat exchanger. This absorption of heat by the system - refrigerant or the working fluid - from the surroundings (ambient or return air) is shown in Figure 8.4 as the shaded area under the isobar, between entropies s1 and s2. In an isobaric process: Δp = 0 and, Q = ΔH The later mathematical statement, Q = ΔH, implies that in this isobaric process, the heat absorbed by the refrigerant, during the evaporation phase, results in a net increase in the enthalpy of the refrigerant.

152  Thermodynamic Processes Some of the equations with practical applications in closed system isobaric processes are listed below: υ  T2 = T1  2   υ1 

(8.5)

T  υ2 = υ1  2   T1 

(8.6)

q = h2 - h1

(8.7)

q = c p (T2 - T1 )

(8.8)

q = cn (T2 - T1 ) + p ( υ2 - υ1 )

(8.9)

W = p ( υ2 - υ1 )

(8.10)

W = R(T2 - T1 )

(8.11)

Where, q = Heat per unit mass or, Q/m = Total Heat/Unit Mass, in BTU/lbm or kJ/kg p = Pressure in lbf/ft2 or Pa υ1 = Initial specific volume in ft3/lbm or m3/kg υ2 = Final specific volume in ft3/lbm or m3/kg h1 = Initial enthalpy in BTU/lbm or kJ/kg h2 = Final enthalpy in BTU/lbm or kJ/kg T1 = Initial Temperature, in °F, °C, °K, °R T2 = Final Temperature, in °F, °C, °K, °R R = Specific gas constant, in ft lbf/lbm °R or kJ/kg °K W = Specific work in ft lbf/lbm or kJ/kg. W also represents total work in ft lbf or kJ Note that Eq. 8.5 and Eq. 8.6 are derived from the ideal gas law, as stated in form of Eq. 8.12 below, with pressure held constant in an isobaric process. p1n1 p2 n2 = T1 T2

(8.12)

8.4.5 Isobaric process example II: Isobaric segments of an ideal cycle heat engine As shown in Figure 8.5, ideal heat engine cycle segments represented by processes A-B and C-D are isobaric processes because during these two

8.4  Adiabatic Process Example II – Compressor Segment of a Refrigeration System  153

Figure 8.5

Isobaric process in an ideal heat engine cycle.

processes within the heat engine cycle, the pressure remains constant as the volume expands and contracts, respectively. The working fluid performs work on the surroundings as it expands, from point A to B. This work is considered positive. The surroundings perform work on the system, or working fluid, from point C to D; resulting in negative work. As explained in greater detail under the heat cycle section, this negative work is in form of the condensate pump performing work on the condensed vapor as it is pressurized to saturated liquid phase. 8.4.6 Constant temperature or isothermal process An isothermal process is a thermodynamic process in which the temperature stays constant. In isothermal processes, there is no change in internal energy because internal energy is directly related to temperature. This is validated by Eq. 8.17. Furthermore, as stipulated by Eq. 8.18, there is no change in enthalpy. Some of the equations with practical applications in closed system isothermal processes are listed below: T2 = T1

or ,

υ  p2 = p1  1   υ2 

ΔT = 0

(8.13) (8.14)

154  Thermodynamic Processes  p  υ2 = υ1  1    p2 

(8.15)

q = W (8.16) u2 - u1 = 0(8.17) h2 - h1 = 0(8.18)  p  Q = nR∗T ln  1  (8.19)  p2 

Where, q = Heat per unit mass or, Q/m = Total Heat/Unit Mass, in BTU/lbm or kJ/kg Q = Total heat in BTU or kJ. Q also denotes molar heat in BTU/ lbmol or kJ/kmol p1 = Initial Pressure in lbf/ft2 or Pa p2 = Final Pressure in lbf/ft2 or Pa υ1 = Initial specific volume in ft3/lbm or m3/kg υ2 = Final specific volume in ft3/lbm or m3/kg h1 = Initial enthalpy in BTU/lbm or kJ/kg h2 = Final enthalpy in BTU/lbm or kJ/kg T1 = Initial Temperature, in °F, °C, °K, ˚R T2 = Final Temperature, in °F, °C, °K, ˚R n = Number of moles, in lbmole or kmol R = Specific gas constant, in ft lbf/lbm °R or kJ/kg °K R* = Universal gas constant, in ft lbf/lbmole °R or kJ/kmol °K W = Specific work in ft lbf/lbm or kJ/kg. W also represents total work in ft lbf or kJ 8.4.7  Isothermal process example I: Steam generation process The, latent, water evaporation stage in the steam generation process is an isothermal process because the temperature of the water and saturated vapor remains constant until all evaporation is concluded. The work “W” performed by, or on, an isothermal system, is shown in Figure 8.6. Where work W = W × → Y . And, WX →Y

VY = - p.dV (8.20) VX

∫

8.4  Adiabatic Process Example II – Compressor Segment of a Refrigeration System  155

Figure 8.6  Isothermal process in a heat engine heat cycle.

Example Problem 8.1 Consider the collision of the cast iron block with the compressed air-filled shock absorbing system described in Case Study 2, part (e), Chapter 1. A cooling jacket is installed on the cylinder to maintain the temperature constant, at 20°C. For simplicity, assume that 1.3 kg of air is present in the shock absorbing system cylinder. Calculate the amount of heat, in BTU’s, the cooling jacket would need to remove each time a block is stopped. Solution According to Eq. 8.19, the heat removed to maintain constant temperature in an isothermal compression process would be:  p  Q = nR∗T ln  1   p2 

Where, Q = Total heat removed n = Number of kmols of gas involved =? R* = Universal Gas Constant = 8.314 kJ/kmol.°K or, 1545 ft-lbf/ lbmole.°R. See Table 8.1 below.

156  Thermodynamic Processes p1 = Initial pressure = 101.3 kPa = 1 Bar p2 = Final Pressure = 202.6 kPa = 2 Bar T = 20°C + 273 = 293°K Since the number of kmols is not given, it needs to be derived using given mass of air, i.e., 1.3 kg and the molecular weight listed in Table 8.1. The molecular weight of air, from Table 8.1 is 28.97 kg/kmol. Therefore, n = Number of kmols of air = 1.3 kg / 28.97 kg/kmol = 0.0449 kmols Then, from Eq. 8.19:  1 Q = (0.0449kmol )(8.314kJ / kmol ° K )(293° K ) ln    2

= -76 kJ of heat removed. Since 1kJ = 0.95 BTU, Qremoved, in BTU’s = (-76 kJ)(0.95 BTU/kJ) = -71.2 BTUs 8.4.8  Constant volume process A constant volume process is also referred to as an isometric process or an iso-volumetric process. In a constant volume thermodynamic process, the volume of a closed system remains constant while other parameters, i.e., pressure, internal energy and temperature may vary. Pressure energy, work, volume, and pressure are related by the following equation: ΔW = PΔV, where P is pressure Since the volume is constant in an isometric process, ΔV = 0, ΔW = 0 Therefore, in an isometric or constant volume process, according to the First Law of Thermodynamics: Q = ΔU In other words, in an isometric or isochoric process, the heat added to the system is transformed into the higher level of the system’s internal energy “U.”

8.4  Adiabatic Process Example II – Compressor Segment of a Refrigeration System  157 Table 8.1  Molecular weights of common gases associated with combustion reactions and byproducts.

Gas Molecula Weight 26.04 Acetylene, C2H2 Air 28.966 Ammonia (R-717) 17.02 Argon, Ar 39.948 58.12 N-Butane, C4H10 Iso-Butane 58.12 Carbon Dioxide, CO2 44.01 Carbon Disulphide 76.13 Carbon Monoxide, CO 28.011 30.07 Ethane, C2H6 Ethyl Alchohol 46.07 Ethylene, C2H4 28.054 Helium, He 4.02 2.016 Hydrogen, H2 Hydrogen Chloride 36.461 Hydrogen Sulfide 34.076 16.044 Methane, CH4 Methyl Alcohol 32.04 Methyl Butane 72.15

Gas Molecular Weight Natural Gas 19 Nitric Oxide, NO2 30.006 28.0134 Nitrogen, N2 Nitrous Oxide 44.012 Oxygen, O2 31.9988 Ozone 47.998 Propane, C3H8 44.097 Propylene 42.08 R-11 137.37 R-12 120.92 R-22 86.48 R-114 170.93 R-123 152.93 R-134a 102.03 R-611 60.05 Sulfur 32.02 Sulfur Dioxide 64.06 Sulfuric Oxide 48.1 Wator Vapor/Steam, 18.02

8.4.9 Constant volume process example I: Superheated steam generation in a “rigid” constant volume boiler Most boilers consist of rigid vessels, tanks, channels or tubes. Since these systems are rigid, as the water is boiled, evaporated into saturated steam, and heated further into superheated steam, the temperature and pressure of the water or steam increase but the volume remains constant. There are two basic approaches to boiler design: (1) Water Tube Boiler and (2) Fire Tube Boiler. The water tube boilers are more common. Figure 8.7 illustrates the fundamental design concept for Water Tube Boilers. 8.4.10  Constant volume process example II: Ideal heat engine Processes or paths D-A and B-C in an ideal heat engine cycle, as shown in Figure 8.5 Pressure–Volume diagram, provide another example of constant volume or isometric thermodynamic process. In paths D-A and B-C, while the pressure increases and decreases, respectively, the volume stays constant.

158  Thermodynamic Processes

Figure 8.7  Fire-tube boiler design and operation concept diagram.

8.4.11  Isentropic or constant entropy process Isentropic process, in a thermodynamic system, is a process in which the entropy of the system stays constant. Any reversible adiabatic process is an isentropic process. 8.4.12 Isentropic process example I: Ideal heat engine – carnot cycle Process paths D′-A′ and B′-C′, as shown in the Temperature-Entropy diagram for a Carnot Cycle Heat Engine in Figure 8.8, are examples of isentropic process. Note: The Carnot Cycle is explained later in this chapter. In paths D′-A′ and B′-C′, while the temperature rises and drops, respectively, the entropy “s” stays constant. 8.4.13  Throttling process and inversion point Throttling process, in a thermodynamic system, is an adiabatic process which consists of a significant pressure drop but no change in the system enthalpy. This significant drop in pressure is sometimes accompanied by significant increase in velocity of the thermodynamics system. Furthermore, in

8.4  Adiabatic Process Example II – Compressor Segment of a Refrigeration System  159

Figure 8.8  Isentropic processes in a carnot cycle, an ideal cycle heat engine.

a throttling process, no heat is exchanged with the surrounding and in some cases no work is performed on or by the system. Since the enthalpy in a throttling process stays constant, a throttling process is also an isenthalpic (constant enthalpy) process. In a throttling process: p2 < mvapor. Note that all points on path 2 – 3 represent wet vapor state. Point 3: This point also lies in the area of the graph where the refrigerant exists in vapor-liquid mixture form. However, point 3 is closer to the saturated vapor line than it is to the saturated liquid line. This means that, at point 3, the percentage of vaporized refrigerant is greater than the percentage of liquid. In other words, mvapor > mliquid. Refrigerant compression process begins at point 3 and extends up to point 4. Point 4: This point lies directly on the saturated vapor line. There is no liquid refrigerant along this line; mliquid = 0 and, therefore, ω = 1, or 100%. The refrigerant is in pure vapor phase all along this line. However, this vapor state is a saturated vapor state. Any loss of heat at this point would slide the refrigerant back into the condensed, or partially condensed, phase. On the other hand, if this point were to shift to the right of the saturated vapor line, it would be in superheated vapor phase. Superheated vapor is also referred to as “dry vapor.” 11.3.3  Dry vapor compression process The compression efficiency in a refrigeration cycle can be enhanced by extending the refrigerant evaporation process all the way to the saturated vapor line or beyond. This is precisely the strategy employed with dry vapor compression process as shown in Figure 11.8. In dry vapor compression refrigeration systems, the refrigerant leaves the evaporator section in either saturated vapor or superheated form as shown by point 3 in Figure 11.8. Furthermore, as obvious from Figure 11.8, the temperature of the refrigerant rises during the compression phase in dry vapor compression systems. As the temperature of the refrigerant rises, the refrigerant vapor becomes superheated. This is affirmed by the location of point 4 in Figure 11.8. Following equations find common application in refrigeration cycles: Computation of refrigerant compressor power utilizing the compression path 3-4: P = W = m(h4 - h3 )(11.2)

Where, Ẇ represents “workflow rate” in BTU/s or J/s. Computation of refrigerant mass flow rate ṁ, utilizing the evaporation path 2-3: m=

Qin (11.3) (h3 - h2 )

242  Refrigeration Cycles and HVAC Systems

Figure 11.8  Dry vapor compression cycle in refrigeration systems.

Where, Qin represents “heat flow rate” in BTU/s or J/s. Rearrangement of Eq. 11.3 yields the equation for heat flow rate calculation: Qin = m(h3 - h2 ) (11.4)

These equations are premised on the refrigeration cycle depicted in Figure 11.8. The nomenclature used in these equations is specific to the refrigeration cycle shown in Figure 11.8. If letters are used to denote various points in the refrigeration cycle, the numerical subscripts are replaced with letters as illustrated in Case Study 11.1 below. 11.3.4 Coefficient of performance, or COP, in refrigeration systems – refrigerator example Coefficient of performance of a refrigeration system – in this case, that of a refrigerator - is defined as the ratio of useful energy transfer to the work input. The thermodynamic system is considered to be the refrigerant in this computation of COP. The equations for the COP are as follows: COPrefrigerator =

Qin (11.5) (Qout - Qin )

11.4  SEER, Seasonal Energy Efficiency Ratio   243

Also, COPrefrigerator =

Qin (11.6) Win

And, COPrefrigerator = COPheatpump operation - 1(11.7)

EER = 3.41 × COP 

(11.8)

11.4  SEER, Seasonal Energy Efficiency Ratio SEER or Seasonal Energy Efficiency Ratio is a rating based on the cooling output in BTU during a cooling season divided by the total electrical energy drawn from the utility, during the same period, in Watt-Hours. Therefore, the engineering units for SEER rating are BTU/W-h. The higher the SEER rating of an air-conditioning system the more efficient it is. Example 11.1 As an Energy Engineer, you have been asked by your client to determine the total annual cost of electrical energy consumed and the input power demanded by an air conditioning system with the following specifications: SEER Rating: 10 BTU/W-h Air Conditioning System Rating: 10,000 BTU/hr Total, Annual, Seasonal Operating Period: 130 days, 8 hours per day. Average, Combined, Electrical Energy Cost Rate: $0.18/kWh Solution Annual Cost of Energy = ($0.18/kWh).(Total Energy Drawn from The Utility, Annually) Total Power Demanded from The Utility = (Air Conditioning System Rating, in BTU/hr)/(SEER Rating, in BTU /W-hr (11.9) Note: Both BTU values in Eq. 11.9 are outputs, while the W-hr value represents the input energy drawn from the line (utility) side of the power distribution system.

244  Refrigeration Cycles and HVAC Systems \Total Power Drawn from The Utility = (10,000 BTU/hr)/(10 BTU /W-hr) = 1,000 Watts, or 1 kW Then, Total Energy Drawn from The Utility, Annually = (1 kW).(Total Annual Operating Hours) = (1 kW).(130 Days).(8 Hours/Day) = 1,040 kWh \Total Annual Cost of Electrical Energy Consumed = (1,040 kWh)/($0.18/kWh) = $187.20

11.5  Case Study 11.1: Refrigeration Cycle An air-conditioning system uses HFC-134a refrigerant. The refrigeration system is cycled between 2.0 MPa and 0.40 MPa. A pressure-enthalpy diagram for HFC-134a is presented on the next page. a. Draw the refrigeration cycle on the given diagram. See Figure 11.9.  b. Determine the change in entropy during the throttling process. c. Determine the percentages of liquid and vapor at the end of the throttling segment of the refrigeration cycle. d. How much enthalpy is absorbed by the system (refrigerant) in the evaporation (latent) phase?  e. How much enthalpy is extracted from the system (refrigerant) in the condensation (condenser) phase of the cycle?   f. In which leg of the refrigeration cycle would expansion be used? g. If the refrigeration capacity of this system were sized based on the enthalpy extracted from the refrigerant, as calculated in part (e), what would the specification be in tons (Metric). (a) Solution - The process involved in the drawing of the refrigeration cycle is as follows: C - D: See Figure 11.9. Locate the 2 MPa and 0.4 MPa points along the pressure (vertical) axis of the chart, name these points “C” and “D,” respectively. This

Figure 11.9  Pressure-enthalpy diagram, Case study 11.1.

11.5  Case Study 11.1: Refrigeration Cycle  245

246  Refrigeration Cycles and HVAC Systems is the throttling portion of the refrigeration cycle. Note: HFC-134a, at point C is in, high pressure, saturated liquid phase. Throttling process is adiabatic and D h = 0 ∴ Draw a straight, vertical, line down from C to D. At point D, R-134a is in liquid-vapor mixture phase. D - A: See Figure 11.9. The next step involves complete transformation of the refrigerant from liquid to gaseous phase through absorption of heat, or D h. This is an, non-adiabatic, isobaric process; so, draw a straight, horizontal, line from D to A. This step is referred to as the evaporator segment of the refrigeration cycle. This is where the system (refrigerant) performs cooling of the environment as its phase undergoes through latent transformation from liquid to gaseous phase. A - B: See Figure 11.9. The next step involves the transformation of HFC-134a from LOW pressure (0.4 MPa) gaseous phase to HIGH (2 MPa) pressure gaseous phase. This is the compressor segment of the refrigeration cycle. This phase is an isentropic process, D s = 0, therefore, draw a straight line from point A to B, asymptotic to S @ 1.73 kJ/kg.°K. B - C: See Figure 11.9. The next step involves the transformation of HFC-134a from high (2.0 MPa) pressure gaseous phase to high pressure, saturated, liquid phase. This segment constitutes the condenser segment of the refrigeration cycle. This is an isobaric process, D P = 0. Therefore, draw a straight line from point B to C, along P = 2 MPa line. b. Determine the change in entropy during the throttling process:

Solution: See Figure 11.9. D s = sD – sC @ 1.35 – 1.33 @ 0.02 kJ/kg °K.

c. Determine the percentages of liquid and vapor at the end of the throttling segment of the refrigeration cycle. Solution This involves reading the value of “x,” the quality, at point “D,” from the Pressure-Enthalpy diagram. See Figure 36. x = (mvapor) / (mvapor + mliquid) = 0.45 or 45%

11.5  Case Study 11.1: Refrigeration Cycle  247

In other words, mvapor (%) = 45 %, And since: (%mvapor + %mliquid) = 100%, mliquid (%) = 100 – 45 = 55% d. How much enthalpy is absorbed by the system (refrigerant) in the evaporation (latent) phase? Solution This involves step D – A, see the Pressure – Enthalpy Diagram, see Figure 36: Dh D – A Phase= hA – hD = 400 – 300 = 100 kJ/kg e How much enthalpy is extracted from the system (refrigerant) in the condensation (condenser) phase of the cycle? Solution This involves step B– C, see Figure 36. : Dh B – C Phase= hB - hC = 440 – 300 = 140 kJ/kg f. In which leg of the refrigeration cycle would expansion occur? Answer: The throttling leg, Step C – D. See Figure 11.9. g. Determination of the refrigeration capacity in tons based on the enthalpy extracted from the refrigerant, as calculated in part (e): Solution Heat extracted from the refrigerant in part (e) = 140 kJ/kg. Rate of refrigeration for this system = (140 kJ/kg).(1000 kg)/24 hr = 140,000 kJ/24 hr = 140,000 kJ/(24 hr)/(3600s/hr) = 1.62 k.J/s = 1.62 kW Since 1 tonne of refrigeration amounts to 3.86 kW, the refrigeration capacity of this system, in tons, would be: = (1.62 kW)/(3.86 kW) = 0.4 tonne.

248  Refrigeration Cycles and HVAC Systems

11.6  Direct Digital Control of HVAC Systems Like many manufacturing operations and chemical process, nowadays, HVAC systems are taking advantage of automation and automated control systems. Automated closed loop control systems permit operator-error-free and reliable operation of HVAC systems. Prior to the 1990’s and 1980’s, proper and effective operation of many large HVAC systems in industrial and commercial domains required a sizeable crew of utilities engineers and technicians. Such modes of operation still exist in large industrial facilities where businesses have not committed capital investments to automate HVAC Systems. In non-automated industrial HVAC operations, the sole responsibility of HVAC technicians, often, is to monitor, track, audit, and optimize the performance of chillers, compressors, air washers, cooling towers, fans, and pumps, constantly. This manual approach of continuous manual monitoring, data recording, and frequent manual adjustments of HVAC controls, involves significant labor cost with limited benefits. Consequently, despite the best effort, in manually monitored and controlled HVAC systems, the operations in production environment occur under dry-bulb and wet-bulb conditions that are outside of the optimal ranges – in many cases, in essence, producing product that does not meet strict product quality control requirements and specifications. The advent of industrially hardened computers and PLC’s, Programmable Logic Controllers have ushered the era of automation in the HVAC realm. Central controllers, whether they consist of PC’s or PLC’s, are programmable, meaning the control system code/program may be customized for the specific use. Major program features, within the overall application program, include synchronous controlled events, time schedules, set-points, control logic, timers, trend logs, regression analysis based forecasts, alarms, graphs, graphical depictions of the HVAC systems with live or real-time data points. There are myriad alternative automated HVAC brands in the market to choose from. Some automated HVAC System Manufacturers/Suppliers and brands are well established and well supported subsidiaries of large firms. End users in the market for automated HVAC systems are advised to consider the following criteria in the formulation of their decision for a specific brand: 1.

Is the automated HVAC system provider well established in the market with reasonable availability of technical support during start-up phase, commissioning phase, and post installation operation?

2.

Ensure that the technology - hardware, software, and firmware - offered by the vendor is well beyond its “infancy” period, and therefore, vetted or proven.

11.6  Direct Digital Control of HVAC Systems  249

3.

If possible, avoid technology that is proprietary and offers little compatibility with the mainstream PC’s and PLC’s.

4.

Select brands and technology that is compatible with established, recognized, standard communication protocol – such as that which is established and sponsored by recognized entities like IEEE, Institute of Electrical and Electronic Engineers.

5.

Choose field input and output hardware that is versatile and compatible with mainstream sensors and output transducers

6.

Preference should be given to PC’s and PLC that operate on - or are compatible with - mainstream operating system platforms, i.e., Microsoft Windows 11, Vista, and equivalent late generation systems. This is to ensure compatibility with useful application software packages, such as the Wonderware â, AKA, Dundas, HMI, Human Machine Interface package and other equivalent software packages.

HVAC control systems are sometimes embedded into comprehensive EMS, Energy Management Systems, or BMS, Building Management Systems. This approach offers the following advantages: 1.

Cost reduction through economies of scale.

2.

More attractive return on investment and economic justification.

3.

Obvious confluence of automated HVAC system projects with other energy productivity improvement projects. This could potentially provide additional advantage of financing such projects through ESCO/ EPC programs. (Reference: Finance and Accounting for Energy Engineers, By S. Bobby Rauf, Fairmont Press)

4.

Central monitoring and control of all utilities in an industrial or commercial facility.

5.

Computerized streamlining and scheduling of HVAC system PM programs along with the PM and Predictive Maintenance of all other plant equipment.

General approach to the architecture of most automated HVAC systems is illustrated in Figure 11.10. As shown in this system architecture, the core brain of the system is the CPU, Central Processing Unit. This CPU can be a PLC, Programmable Logic Controller, a DDC, Direct Digital Controller, or a simply an industrially hardened PC. The language or code utilized to program the PLC, or the PC would be proprietary and specific to the type

Figure 11.10  Automated HVAC control system architecture.

250  Refrigeration Cycles and HVAC Systems

11.6  Direct Digital Control of HVAC Systems  251

of PLC or PC installed. Specifications such as the size of CPU memory, RAM, Random Access Memory and the ROM, Read Only Memory, would determine the length of code or program that can be written and the number of data points that can be monitored, tracked, trended, and controlled. The heart of the CPU is referred to as a “microprocessor.” Controllers or CPU’s that drive large HVAC systems, or combination of large HVAC and EMS Systems, are sometimes equipped with dual microprocessors to limit the program scan times. Program scan times are sometimes also referred to as cycle times. Reasonable scan time for an average automated HVAC system is approximately 30 milliseconds. Shorter scan times are desirable. Longer cycle times can, potentially, cause the control system to become ineffective and dysfunctional. As shown in Figure 11.10, the CPU is sometimes connected to a Monitoring Terminal, which includes a PC, display/monitor, keyboard, and other suitable peripheral devices. This PC System serves as an interface between the CPU and the programmers, maintenance technicians and certain qualified production personnel. Often, this is where the control program and application software reside. The PC and the monitor shown at the bottom of Figure 11.10 serve as the process annunciation system, complete with HVAC graphics, real time data, alarms, trend charts, event logs, and production performance data diagnostics. This system, or sub-system, is referred to as the Human Machine Interface or HMI System. This PC based system is equipped with a suitable operating system such as the Microsoft Windows, or Vista, and an HMI software package, such as, Wonderware, or equivalent. Note that around the time of publication of the 2nd edition of this text, in 2022, Wonderware solutions had transitioned to AVEVA Solutions. The HMI system and/or the CPU Terminal are often connected, in network format, to other IT, Information Technology, and accounting computers for monitoring of productivity and production cost tracking purposes. As shown in the automated HVAC system architecture diagram, the HMI system and the CPU terminal are sometimes linked to remote or off-site locations, i.e., corporate offices, through Ethernet, wireless routers, and modems. As shown in Figure 11.10, automated HVAC systems also consist of other peripheral equipment, such as, input and output modules. Input modules receive different types of signals from peripheral devices/sensors and process them for presentation to CPU. Output modules take various outputs or commands from the CPU and package them such that they can be used to either turn on or turn off control equipment or to gradually

252  Refrigeration Cycles and HVAC Systems increase or decrease magnitudes of various parameters to maintain HVAC control. Inputs and outputs, in essence, are signals or commands. Main categories of inputs and outputs are as follows: 1.

Digital or discrete inputs

2.

Digital or discrete outputs

3.

Analog inputs

4.

Analog outputs

11.6.1  Digital or discrete inputs The digital inputs - sometimes referred to as discrete inputs - are simply closed or open contact signals that represent the closed or open status of switches in the field. These switches or contacts can be limit switches, safety interlock switches or auxiliary contacts of motors, sensors, etc. The normally closed safety interlock contact shown in Figure 11.11 presents 110 volts AC to the discrete input block. Closed contact type inputs can present other “nonzero” voltages, such as, 5 volts DC, 10 volts DC and 24 volts AC/DC, or 110 volts AC to the discrete input modules. The end result is a logic level HIGH or “1” to the CPU for computation and decision making purposes. On the other hand, an open contact type input, such as the one shown to represent the normally open pressure switch in Figure 11.11, would be interpreted as a LOW logic level or “0” by the discrete input module and the CPU. Another perspective on the role of input modules would be to view them as intermediary devices that transform and isolate discrete signals being received from peripheral devices. 11.6.2  Digital or discrete outputs Digital outputs are typically generated by relay contacts based in the discrete output modules. A “1” or a “HIGH” from the CPU is transformed into a 5 volts DC, 10 volts DC, 24 volts AC/DC, or 110 volts AC signal, which can then be used to start and stop equipment, as a part of the overall automated HVAC control scheme. For example, a “1” from the CPU is transformed into a 110 volt AC signal by the 110 Volt AC Discrete Output Module in Figure 11.11. This 110 Volt AC output from the output module is fed to the AC motor starter coil to turn on the motor, in response to the specific input conditions and the program or algorithm. Another discrete output from the 110

Figure 11.11  Automated HVAC control system architecture.

11.6  Direct Digital Control of HVAC Systems  253

254  Refrigeration Cycles and HVAC Systems Volt AC Discrete Output Module, in Figure 11.11, is fed to an alarm horn to annunciate an alarm condition. Note that in both examples the second terminal of each controlled field device is connected to the power system neutral, designated as “N.” 11.6.3  Analog inputs Analog inputs represent gradually varying parameters or signals. Analog signals, in automated HVAC systems, include temperature, humidity, volume, pressure and even electrical current drawn by fan, blower or pump motors. Common analog current range is 4 – 20 mA. In certain cases, signals representing pressure, temperature, and volume are converted into analog voltage signals. These analog voltage signals range, commonly, from 0 – 10 Volts DC. Analog input modules convert signals from thermocouples, RTD’s, Resistance Temperature Detectors, pressures sources and other types of analog signal sources, into digital logic numbers for use by the CPU. See the RTD application example shown in Figure 11.11. The analog temperature measurement from the RTD, in form of low voltage signal, is fed to the Analog Input Module. The analog module converts this voltage signal into an equivalent digital value. This digital value is presented to the CPU for computation or algorithm execution purposes. 11.6.4  Analog outputs Operation of analog outputs can, essentially, be explained as the operation of analog inputs in reverse. Analog output begins with a presentation of a digital or binary number by the CPU to the analog output module. This digital or binary number is then transformed into an equivalent analog signal such as 0 – 5 Volt DC, 0 – 10 Volt DC, 0 – 110 Volt AC or 4 – 20 mA DC signal. Any of such analog signals can then be used to command the gradual or analog operation of HVAC control equipment like valves, variable frequency drives, heat sources, electric valve actuators, pneumatic actuators, etc. In Figure 11.11, the Analog Output Module is shown feeding a 4 – 20 mA analog output to a VFD, Variable Frequency Drive. The drive can base its output power frequency on this 4 – 20 mA analog input to control the speed of an HVAC fan or pump motor.

Figure 11.12  Pressure-enthalpy diagram, HFC-134a.

11.6  Direct Digital Control of HVAC Systems  255

256  Refrigeration Cycles and HVAC Systems

Chapter 11 Self-Assessment Problems and Questions 1.

An air-conditioning system uses HFC-134a refrigerant. The pressure-enthalpy diagram for this refrigerant is presented on the next page. The refrigeration system is cycled between 290 psia and 60 psia.



a. Draw the refrigeration cycle on the given diagram 



b. What is the change in enthalpy during the expansion process? 



c. Determine the percentages of liquid and vapor at the end of the throttling segment of the refrigeration cycle.



d. How much enthalpy is absorbed by the system (refrigerant) in the evaporation (latent) phase? 



e. How much enthalpy is extracted from the system (refrigerant) in the condensation (condenser) phase of the cycle?  



f. Determine the percentages of liquid and vapor at B.



g. Assume that the mass flow rate of refrigerant being cycled in this air-conditioning system is 10 lbm/min and the compressor efficiency is 70%. Determine the amount of electrical power demanded by the compressor motor if the compressor motor efficiency is 90%.



h. Which leg of the refrigeration cycle would be considered isentropic?

I. A-B II. B-C III. C-D IV. D-A

2.

As stated in this chapter, 1 tonne (SI/Metric) of refrigeration capacity is equivalent to 3.86 kW of power. Provide the mathematical proof for this equivalence.

3.

As an Energy Engineer, you are performing an energy cost assessment for operating a 20,000 BTU/hr air conditioner. Based on the data and specifications provided below, determine the total annual cost of

Self-Assessment Problems and Questions   257

electrical energy consumed and the input power demanded by the air conditioning system:

•• SEER Rating: 12 BTU/W-h •• Air Conditioning System Rating: 20,000 BTU/hr •• Total, Annual, Seasonal Operating Period: 200

days, 10 hours

per day.

•• Average, Combined, Electrical Energy Cost Rate: $0.20/kWh

Appendices Appendix A

This appendix includes the solutions and answers to end of chapter self-assessment problems and questions

Chapter 1—Self-Assessment Problems and Questions 1.

Determine the amount of heat extracted by the quench water, per block, in Case Study 1.2, using the temperature rise of the water when the steel block is dropped into the quenching tank. The temperature of the block is 100°C when it enters the quench water. The initial temperature of the water in the quench tank is 20°C and volume of water is 6.038 m3. The final, equilibrium, temperature of the water and the block is 30°C.

Note: This is an alternate approach to the solution for part (g) in Case Study 1.2. This solution is premised on the use of the rise in water temperature to calculate the amount of heat extracted by the quench water, per block. Solution Given or known : = 4.186 kJ/kg °K  c water = (1000 kg/m3)(6.038 m3) mwater = 6,038 kg Twater - i = 20°C = 273 + 20°C = 293°K Twater - f = 30°C = 273 + 30°C = 303°K ∴ ΔTwater = 303°K - 293°K = + 10°K Since Q absorbed by water Q absorbed by water Q absorbed by water ∴ Q absorbed by water   - Q lost by the block

{From Table 1.5}

= (m water)(c )(ΔTwater) = (6,038 kg)(4.186 kJ/kg °K )(+10°K) = 252,750 kJ = = 252,750 kJ @ 252,770 kJ 259

260  Appendix A

Chapter 2—Self-Assessment Problems and Questions 1.

As an engineer, you are charged with the task to estimate the amount of electrical power produced, in MW, by a steam based power generating plant. Assume that there is no heat loss in the turbine system and that difference between the enthalpies on the entrance and exit ends of the turbine is completely converted into work, minus the inefficiency of the turbine. All of the data available, pertinent to this project, is listed below:

•• Electrical Power Generator Efficiency, η r: 87% •• Steam Turbine Efficiency, η : 67% •• Mass flow rate for steam, ṁ: 20 kg/s (44 lbm/s) •• Exit enthalpy, h , of the steam: 2900 kJ/kg (1249 BTU/lbm) •• Incoming superheated steam enthalpy, h : 3586 kJ/kg (1545 BTU/ Generato

Turbine

f

lbm)

i

Solution Solution Strategy: As explained in Chapter 2, the power delivered by steam to the turbine blades, Psteam, in a simplified, no heat loss, no kinetic head loss, no potential head loss and zero frictional head loss scenario can be represented by the mathematical relationship stated in form of Eq. 2.1. And, in the context of flow of energy from steam to electricity, functional relationship between electrical power, PElectrical, generator efficiency ηGener, steam turbine efficiency ηTurbine, and Psteam can be expressed in form of ator Eq. 2.2. Psteam = (hi - hf)ṁ(2.1) PElectrical = (Psteam ) * (ηTurbine) * (ηGenerator)(2.2) Solution in SI/Metric Units The power imparted by the steam onto the turbine blades, Psteam, can be determined by applying Eq. 2.1: Psteam = (3586 kJ/kg - 2900 kJ/kg)(20 kg/s) = 13,720 kJ/s

Chapter 2—Self-Assessment Problems and Questions   261

Then, the electrical power produced by generator, PElectrical, can be determined by applying Eq. 2.2: PElectrical = (Psteam ) * (ηTurbine) * (ηGenerator) PElectrical = (13,720 kJ/s) * (0.67) * (0.87) = 7,997 kJ/s Since 1.00 kJ/s = 1 kW, PElectrical in kJ/s = (7,997 kJ/s)/(1.00 kJ/s/kW) = 7,997 kW Since 1,000 kW = 1 MW PElectrical in MW = (7,997 kW)/(1,000 kW/MW) = 8 MW Solution in US/Imperial Units The power imparted by the steam onto the turbine blades, Psteam, can be determined by applying Eq. 2.1: Psteam = (1545 BTU/lbm - 1249 BTU/lbm)(44 lbm/s) = 13,024 BTU/s Then, the electrical power produced by generator, PElectrical, can be determined by applying Eq. 2.2: PElectrical = (Psteam ) * (ηTurbine) * (ηGenerator) PElectrical = (13,024 BTU/s ) * (0.67) * (0.87) = 7,592 BTU/s Since 1.055 kJ/s = 1BTU/s, PElectrical in kJ/s = (7,592 BTU/s)(1.055 kJ/s) = 8010 kJ/s Since 1 kJ/s = 1 kW PElectrical in kW = 8,010 kW Since 1 MW = 1000 kW, PElectrical in MW = (8,010 kW)/(1000 kW/MW) = 8 MW 2.

Consider the scenario described in Problem (1). Your client has informed you that the power generating plant output requirement has now doubled. Based on the concepts and principles learned in Chapter 2, what is the most suitable alternative for doubling the power output if the exit

262  Appendix A enthalpy, hf, of the steam must be kept constant at the original 2900 kJ/kg (1249 BTU/lbm) level?

A. Double the mass flow rate, ṁ, only.



B. Double the incoming superheated steam enthalpy, hi only.



C. Double the efficiency of the turbine.



D. Double the efficiency of the generator.



E. Increase mass flow rate, ṁ, incoming superheated steam enthalpy, hi and increase the efficiency specification on the turbine.

Answer As apparent from inspection of Eq. 2.1 and Eq. 2.2, theoretically, the power output can be doubled by doubling the mass flow rate, ṁ. However, doubling the mass flow rate is not practical under normal circumstances. Psteam = (hi - hf)ṁ(2.1) PElectrical = (Psteam ) * (ηTurbine) * (ηGenerator)

(2.2)

It is also apparent from examination of Eq. 2.1 that the relationship between hi and PElectrical or Psteam is not linear. Therefore, doubling hi is not only impractical from superheating capacity point of view, but could escalate power to a level that is more than twice the original level. Doubling the efficiency of the turbine or the generator is not possible since that would exceed 100% in each case; and efficiency cannot exceed 100%. Hence, the most practical and reasonable answer would be “E.” A suitable combination of an increase in mass flow rate, ṁ, a commensurate increase in incoming superheated steam enthalpy, hi, and increase in the efficiency of the turbine would be optimum and more practical.

Chapter 3—Self-Assessment Problems and Questions 1.

Calculate the volume 1 kg of vapor would occupy under the following conditions: h = 2734 kJ u = 2550 kJ p = 365.64 kPa = 365.64 kN/m2 V=?

Chapter 3—Self-Assessment Problems and Questions   263

Solution By definition, the volume that 1 kg of vapor occupies is the specific volume of the vapor. Specific volume, as defined in detail later in this text, is volume per unit mass. In this case, specific volume would be volume (in m3) on per kg basis. Apply Eq. 3.2, re-arrange it and substitute the given values: h = u + pV  V = (h – u)/p 

(3.1) (3.1a)

V = (2734 kJ – 2550 kJ)/365.64 kN/m2 V = (2734 kNm – 2550 kNm)/ 365.64 kN/m2 = 0.503 m3 Therefore, each kg of vapor would occupy 0.503 m3 under the conditions stated above. 2.

In a certain solar system there are four (4) planets oriented in space as shown in Figure 3.1. As apparent from the orientation of these planets, they are exposed to each other such that heat transfer can occur freely through radiation. All four (4) planets are assumed to be massive enough to allow for the interplanetary heat transfer to be an isothermal phenomenon for each of the planets. Perform all computation in the US Unit System.



a. If the 1,300 BTU/lbm of radiated heat transfer occurs from planet × to planet Y, what would be the entropy changes at each of the two planets?



b. If a certain radiated heat transfer between Planets Y and Z causes an entropy change of -2.9 BTU/lbm °R at Planet Y and an entropy change of 3.1 BTU/lbm °R at Planet Z, what would be the overall, resultant, entropy of this planetary system?

Solution (a) In an isothermal (constant temperature) process, the entropy production, Δs, is a function of the energy transfer rate and the change in the enthalpy, heat transferred and the absolute temperature of the object in question is governed by Eq. 3.5: Δs = q / T abs(3.5) ∴ ΔsX = (-1300 BTU/lbm)/(522°R) = - 2.49 BTU/lbm °R {Due to heat loss by Planet X}

264  Appendix A

Figure A-3.1  Entropy.

And,

ΔsY = (+1,300 BTU/lbm)/(504°R) = +2.5794 BTU/lbm °R {Due to heat gain by Planet Y}

Solution (b): According to Eq. 3.4: Overall Δs Planetary System = ∑ (Δsi ) In this case, Δs Planetary System = ΔsX + ΔsY + ΔsYZ + ΔsZ Or,

3.

 2.49 BTU/lbm °R +2.5794 BTU/lbm °R Δs Planetary System = – 2.9 BTU/lbm °R + 3.1 BTU/lbm °R ∴ Overall Δs Planetary System = 0.29 BTU/lbm °R If the mass of vapor under consideration in problem 1 were tripled to 3 kg, what would be the impact of such a change on the volume?

Answer: Since the final value for volume is on per unit mass basis, if the mass is tripled, the volume would increase three (3) fold, as well, if all other parameters stay constant. This is obvious from inspection of Eq. 3.1a V = (h – u)/p 

(3.1a)

Chapter 4—Self-Assessment Problems and Questions  265

4.

Would Eq. 3.2 be suitable for calculation of enthalpy if all available data is in SI (Metric) units?

Answer: The answer is NO. One ostensible reason why Eq. 3.2 cannot be used for calculation of enthalpy in the SI unit realm is the presence of “J,” the Joules constant. The J has a value of 778 ft-lbf/BTU; which is unit conversion factor in the US Unit system. Therefore, Eq. 3.2 pertains to computation of enthalpy in the US unit realm.

Chapter 4—Self-Assessment Problems and Questions 1.

Using the Mollier diagram, find the entropy of steam at 400°C and 1 Atm.

Solution: See the Mollier diagram in Figure A-4.1. Identify the point of intersection of the 400°C line (or 400°C isotherm) and the constant pressure line (or isobar) of 1 bar. This point of intersection of the two lines is labeled B in Figure A-4.1. Entropy Determination: To determine the entropy at point B, draw a straight vertical line from point B to the bottom, until it intersects with the entropy line. The vertical line intersects the entropy line at, approximately, 8.55kJ/kg.°K. Therefore, sB, or entropy at point B, is 8.55kJ/kg °K. 2.

Heat is removed from a thermodynamic system such that the temperature drops from 450°C, at 1 Atm to 150°C, at 1 Atm. Determine the following:



a. The new, or final, Enthalpy



b. The new entropy



c. The state of steam at 150°C and 1 Atm?

Solution: a) Locate the new point, 150°C, 1 Atm, on the Mollier diagram. This new point is shown as point C in Figure A-4.1. A horizontal line drawn to the left from point C intersects the enthalpy line at approximately 2770 kJ/kg.

266  Appendix A

Figure A-4.1  Mollier diagram, SI/Metric units.

Therefore, the new, or final enthalpy, at 150°C and 1 Atm, is 2770 kJ/kg. b) A vertical line drawn from point C down to the entropy axis intersects the entropy line at approximately 7.7 kJ/kg °K. Therefore, the new or final entropy, at 150°C and 1 Atm, is 7.7 kJ/kg °K .

Chapter 5—Self-Assessment Problems and Questions  267

c) As shown in Figure A-4.1, point C lies above the saturation line, in the region labeled superheated steam. Therefore, at 150°C and 1 Atm, the steam is superheated.

Chapter 5—Self-Assessment Problems and Questions 1.

Using the saturated liquid enthalpy value for hL and the saturated vapor enthalpy value for hV, at 0.2 MPa and 120.2°C, as listed in the saturated steam tables in Appendix B, calculated the value for hfg.

Solution As stated in Eq. 5.1: hfg = hV - hL As read from Table A-5.1: = 2706.2 kJ/kg, and hV = 504.68 kJ/kg hL \ hfg = hV - hL = 2706.2 kJ/kg - 504.68 kJ/kg = 2201.5 kJ/kg The value for hfg, at 0.2 MPa and 120.2°C, as listed in Table A-5.1, is 2201.9 kJ/kg. 2.

Calculate the enthalpy of 450 psia and 970°F superheated steam.

Solution As you examine the superheated steam tables for these parameters, in Appendix B, you realize that exact match for this data is not available in the table. See Table A-5.2 below, for excerpts from the superheated steam tables in Appendix B. While the given pressure of 450 psia is listed, the stated temperature of 970°F is not listed. Therefore, the enthalpy for 450 psia and 970°F superheated steam must be derived by applying interpolation to the enthalpy data listed in the tables for 900°F and 1,000°F. The formula for single interpolation, applied between the stated or available enthalpy values for 900°F and 1000°F, at 470 psia, is as follows: h 970°F, 450 psia = ((h1000°F, 450 psia – h900°F, 450 psia )/(1000°F - 900°F))(970-900) + h 900°F, 450 psia

268  Appendix A

Table A-5.1  Properties of saturated steam, by pressure, SI units.

Properties of Saturated Steam by Pressure Metric/SI Unitss Specific Volume m3/kg

Abs. Press. MPa

Temp°C

0.010

 45.81

0.0010103

0.10

 99.61

0.0010431

0.20

120.21

1.00

179.89

Enthalpy kJ/kg

Sat. Liquid Sat. Vapor Sat. Liquid VL VV hL 14.671

Evap. hfg

Entropy kJ/kg

Abs. Sat. Vapor Sat. Liquid Sat. Vapor Press. hV SL SV MPa

191.81

2392.8

2583.9

0.6492

8.1489

0.010

1.6940

417.44

2258.0

2674.9

1.3026

7.3588

0.10

0.0010.605

0.88574

504.68

2201.9

2706.2

1.5301

7.1269

0.20

0.0011272

0.19435

762.68

2015.3

2777.1

2.1384

6.5850

1.0

Chapter 5—Self-Assessment Problems and Questions  269

By substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table A-5.2: h 950°F, 450 psia = ((1552.4 BTU/lbm – 1468.6 BTU/lbm )/ (1000°F - 900°F))(970 - 900) + 1468.6 BTU/lbm = 1527.3 BTU/lbm Note: The available enthalpy values are circled in Table A-5.2. Table A-5.2  Superheated steam table excerpt, US/Imperial units.

Properties of Superheated Steam US/Imperial Units Abs. Press. psia

Temp. °F 

(Sat. Temp. °F )

500

Note: n is in ft3/lbm, h is in BTU/lbm and s is in BTU/(lbm °R) 700

900

1000

260

n

2.062

2.5818

3.0683

3.3065

(404.45)

h

1262.5

1370.8

1475.2

1527.8

 

s

1.5901

1.6928

1.7758

1.8132

360

n

1.446

1.8429

2.2028

2.3774

(434.43)

h

1250.6

1365.2

1471.7

1525

 

s

1.5446

1.6533

1.7381

1.7758

450

n

1.1232

1.4584

1.7526

1.8942

(456.32)

h

1238.9

1360

1468.6

1522.4

 

s

1.5103

1.6253

1.7117

1.7499

600

n

 

 

1.3023

1.411

(486.25)

h

 

 

1463.2

1518

 

s

 

 

1.577

1.7159

 

 

 

 

 

 

270  Appendix A 3.

Determine the enthalpy of saturated water at 50°C and 1 Bar.

Solution The saturation temperature at 1 Bar, 1 Atm, or 101 kPa, as stated in the saturated steam tables in Appendix B, is 99.6°C or, approximately, 100°C. The saturated water in this problem is at 50°C; clearly below the saturation temperature. Therefore, the water is in a subcooled state. In the subcooled state, saturated water’s enthalpy is determined by its temperature and not the pressure. Hence, the enthalpy of saturated water at 50°C and 1 Bar must be retrieved from the temperature based saturated steam tables. From Appendix B, and as circled in Table A-5.3, the enthalpy of saturated water at 50°C is 209.34 kJ/kg1. Note: Since the water is referred to as “saturated water” and is clearly identified to be subcooled, the enthalpy value selected from the tables is hL and not hV.

1

4.

Determine the enthalpy and specific volume for 14 psia steam with a quality of 65%.

Solution Given: Quality, × = 0.65 Absolute Pressure = 14 psia From saturated steam tables in Appendix B, and the excerpt in Table A-5.4, the values of enthalpies and specific volumes, at 14 psia, are: hL = 177.68 BTU/lbm hV = 1149.4 BTU/lbm nL = 0.01669 ft3/lbm nv = 28.048 ft3/lbm Apply equations 5.2 and 5.5: hx = (1-x)hL + xhV(5.2) nx = (1-x)nL + × nv(5.5) Then, hx = (1- 0.65)(177.68 BTU/lbm) + (0.65)(1149.4 BTU/lbm) hx = 809.30 BTU/lbm And,

nx = (1 - 0.65)(0.01669) + (0.65)(28.048 ft3/lbm) nx = 18.24 ft3/lbm

Table A-5.3  Properties of saturated steam, by temperature, SI units.

Specific Volume m3/kg

Enthalpy kJ/kg

Entropy kJ/kg

Temp. °C

Abs. Press. Sat. Liquid Sat. Vapor Sat. Liquid VL VV hL MPa

Evap. hfg

Sat. Vapor Sat. Liquid Sat. Vapor Temp. hV SL SV °C

20

0.002339

0.0010018

57.7610

83.920

2454.1

2537.5

0.2965

8.6661

20

50

0.012351

0.0010121

12.0280

209.34

2382.7

2591.3

0.7038

8.0749

50

100

0.101420

0.0010435

1.6719

419.10

2257.0

2675.6

1.3070

7.3541

100

200

1.554700

0.0011565

0.1272

852.39

1940.7

2792.1

2.3308

6.4303

200

Chapter 5—Self-Assessment Problems and Questions  271

Properties of Saturated Steam by Temperature Metric/SI Unitss

272  Appendix A

Table A-5.4  Properties of saturated steam, by pressure, US units.

Properties of Saturated Steam by Pressure US/Imperial Units Specific Volume m3/kg

Abs. Press. psia

Temp. °F

1.0

101.69

0.016137

40.

152.91

0.016356

14.0

209.52

0.016697

100

327.82

0.017736

Enthalpy kJ/kg

Sat. Liquid Sat. Vapor Sat. Liquid VL VV hL 333.51

Evap. hfg

Entropy kJ/kg

Abs. Sat. Vapor Sat. Liquid Sat. Vapor Press. hV SL SV psia

69.728

1036

1105.4

0.1326

1.9776

1.0

90.628

120.89

1006.4

1126.9

0.2198

1.8621

4.0

28.048

177.68

972.0

1149.4

0.3084

1.7605

14.0

298.57

889.2

1187.5

0.4744

1.6032

4.4324

100

Chapter 6—Self-Assessment Problems and Questions   273

Chapter 6—Self-Assessment Problems and Questions 1.

A boiler is relocated from sea level to a location that is at an elevation of 10,000 ft MSL. Using the table below and Table A-6.2, determine the temperature at which the water will boil if the boiler is assumed to be open to atmosphere. Table A-6.1

Altitude With Mean Sea Level as Ref.

Absolute Pressure in Hg Column Inches Hg mm Hg Column Column 29.9 765 29.4 751 28.9 738 28.3 724 27.8 711 27.3 698 26.8 686 26.3 673 25.8 661 25.4 649 24.9 637 24 613 23.1 590 22.2 568 21.4 547

Feet 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 6000 7000 8000 9000

Meters 0 152 305 457 610 762 914 1067 1219 1372 1524 1829 2134 2438 2743

10000

3048

20.6

15000 20000

4572 6096

16.9 13.8

Absolute Atmospheric Pressure psia 14.7 14.4 14.2 13.9 13.7 13.4 13.2 12.9 12.7 12.5 12.2 11.8 11.3 10.9 10.5

kg/cm2 1.03 1.01 0.997 0.979 0.961 0.943 0.926 0.909 0.893 0.876 0.86 0.828 0.797 0.768 0.739

kPa 101 99.5 97.7 96 94.2 92.5 90.8 89.1 87.5 85.9 84.3 81.2 78.2 75.3 72.4

526

10.1

0.711

69.7

432 352

 8.29  6.75

0.583 0.475

57.2 46.6

Solution/Answer In order to determine the boiling point of the water, at an altitude of 10,000 ft, we need to assess the pressure at that altitude. The pressures at a range of altitudes are available through the given table. In US units, the pressure at 10,000 ft is 10.1 psia, or simply, 10 psia. The next step is to retrieve the saturation temperature at 10 psia, from the saturated steam tables. An excerpt of the Saturated Steam Tables, for the range of pressures surrounding 10 psia, is shown below in form of Table A-6.2.

274  Appendix A

Table A-6.2  Properties of saturated steam by pressure, US units.

Properties of Saturated Steam by Pressure US/Imperial Units Abs. Press. psia

Specific Volume ft3/lbm Temp. in °F

Entropy Btu/(lbm °R)

Enthalpy Btu/lbm

Sat. Liquid Sat. Vapor Sat. Liquid nL nV hL

Evap. hfg

Abs. Sat. Vapor Sat. Liquid Sat. Vapor Press. hV SL SV psia

1036

1105

0.133

1.978

1.0

120.8

1006

1127

0.220

1.862

4.0

38.42

161.2

982

1143

0.2836

1.788

10.0

0.0167

28.0

177.6

972

1149

0.308

1.761

14.0

0.0177

4.4

298.5

889

1188

0.474

1.603

1.0

102

0.0161

333.

4.0

153

0.0164

90.0

10.0

193

0.0166

14.0

209

100

328

69.73

100

Chapter 6—Self-Assessment Problems and Questions   275

Reading directly across, to the right, from the 10 psia saturation pressure field, we see the listed saturation temperature of 193°F. Answer: Therefore, the boiling point for the water at the listed altitude of 10,000 ft would be 193°F. 2.

In problem (1), if the objective is just to heat the water close to the boiling point, will the boiler consume more or less fuel than it did when it was located at the sea level?

Solution/Answer: In order to assess if it would take more heat or less heat to bring the water to a boil at the higher altitude of 10,000 ft, we need to compare the difference between the enthalpy values of water, in saturated liquid phase, at the two different altitudes and pressures. In other words, we need to compare the enthalpy, hL, at 14 psia (pressure at sea level) and 10 psia (pressure at 10,000 ft altitude). Based on the saturated steam table excerpt in Table A-6.2: hl, at 10 psia = 161.2 BTU/lbm, and hl, at 14 psia = 177.6 BTU/lbm Answer: As apparent from the heat content values or enthalpy values above, the water that comes to a boil at the higher altitude of 10,000 ft will do so with less heat content, or would require less heat. Conversely, water that comes to a boil at higher pressure would do so at higher enthalpy value, or greater heat content. 3.

Answer the following questions for water at a temperature of 193°F and pressure of 10 psia:



a. Heat content for saturated water.



b. Specific heat (BTU/lbm) required to evaporate the water.



c. If the water were evaporated, what would the saturated vapor heat content be?



d. What state or phase would the water be in at the stated temperature and pressure?



e. What would the entropy of the water be while it is in saturated liquid phase?

276  Appendix A

f. What would the specific volume of the water be while it is in saturated vapor phase?



g. What would the phase of the water if the pressure is increased to 20 psia while keeping the temperature constant at 193°F?

Solution/Answer a. At193°F and pressure of 10 psia, the water is in saturated liquid form. According to saturated steam table excerpt in Table A-6.2, the saturated water enthalpy at 193°F and pressure of 10 psia is 161.2 BTU/lbm. This value is listed under column labeled hL, in Table A-6.2, in the row representing temperature of193°F and pressure of 10 psia.

Therefore, the enthalpy or heat content for saturated water, at the given temperature and pressure, is 161.2 BTU/lbm.

b. The specific heat, in BTU/lbm, required to evaporate the water from saturated liquid phase to saturated vapor phase, is represented by the term hfg. The value of hfg, for saturated water at 193°F and a pressure of 10 psia, as read from Table A-6.2, is 986 BTU/lbm. See circled values in Table A-6.2. Answer: hfg at193°F and 10 psia = 986 BTU/lbm c Saturated vapor heat content if the water were evaporated would be the value for hv at193°F and 10 psia, and from Table A-6.2 this value is 1143 BTU/ lbm. d. The water would be in saturated liquid phase at the stated temperature and pressure. All stated saturation temperatures and pressures, in the saturated steam tables, represent the current state of water in saturated liquid phase. e. The entropy of water at 193°F and a pressure of 10 psia, in saturated liquid phase, as read from Table A-6.2 would be sL= 0.2836 BTU/ (lbm °R).

Note that the sL value is retrieved form the table and not the sV value. This is because the problem statement specifies the liquid phase.

f. The specific volume of water at 193°F and a pressure of 10 psia, in saturated vapor phase, as read from Table A-6.2 would be nV = 38.42 ft3/lbm. Note that the nV value is retrieved form the table and not the nL value. This is because the problem statement specifies the vapor phase.

Chapter 7—Self-Assessment Problems and Questions   277

g. The phase of the water if the pressure is increased to 20 psia while keeping the temperature constant at 193°F can be determined through the saturated team table excerpt in Table A-6.2. As established earlier in part (d), the current phase of the water is saturated liquid. If, however, the pressure is doubled to 20 psia, according to Table A-6.2, it would take well over 209°F - and additional heat - to reinstate the water into the saturated water phase. Therefore, the water at 20 psia and 193°F would be in subcooled phase or state.

Chapter 7—Self-Assessment Problems and Questions 1.

Why is the efficiency of this power plant in Case Study 7.1 rather low (17%)?

Solution/Answer The efficiency of the power station is low (17%) because of the fact that the working fluid is introduced into the system as -10°C ice. If the working fluid were room temperature water – or return condensate from the discharge side of the turbine, as is the case in Self-Assessment problem number 4 – the overall system efficiency would be substantially higher. 2.

Using the steam tables in Appendix B and the Double Interpolation Method described in Case Study 7.1, US Unit Version, determine the exact enthalpy of a superheated steam at a pressure of 400 psia and temperature of 950°F.

Solution As apparent from the superheated steam tables in Appendix B, the enthalpy value for 400 psia and 950°F is not readily available and, therefore, double interpolation must be conducted between the enthalpy values given for 360 psia, 900°F, and 450 psia, 1000°F, to derive h 400 psia and 950°F. The double interpolation approach, as applied here, will entail three steps. First step involves determination of h400 psia and 900°F, the enthalpy value at 400 psia and 900°F. The enthalpy values available and used in this interpolation step are circled in Table A-7.1. The following formula sums up the mathematical approach to this first step: h400 psia and 900°F    = ((h360 psia and 900°F - h450 psia and 900°F)/(450 psia – 360 psia)). (450 psia - 400 psia) + h450 psia and 900°F

278  Appendix A Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table A-7.1 below: h400 psia and 900°F = ((1471.7 - 1468.6)/(450 psia - 360 psia)) (450 psia - 400 psia) + 1468.6 = 1470.32 BTU/lbm Second step involves determination of h400 psia and 1000°F, the enthalpy value at 400 psia and 1000°F. The enthalpy values available and used in this interpolation step are circled in Table A-7.1. The following formula sums up the mathematical approach to this first step: h400 psia and 1000°F = ((h360 psia and 1000°F - h450 psia and 1000°F)/(450 psia – 360 psia)). (450 psia - 400 psia) + h450 psia and 1000°F Substituting enthalpy values and other given data from superheated steam table excerpt, shown in Table A-7.1 below: Table A-7.1  Superheated steam table excerpt, US/Imperial units.

Properties of Superheated Steam US/Imperial Units Note: n is in ft3/lbm, h is in BTU/lbm and s is in BTU/(lbm °R) Temp. °F

Abs. Press. psia (Sat. Temp.°F ) 260 (404.45)   360 (434.43)   450 (456.32)   600 (486.25)  

n h s n h s n h s n h s

500 2.062 1262.5

700 2.5818 1370.8

900 3.0683 1475.2

1000 3.3065 1527.8

1.5901 1.446 1250.6 1.5446 1.1232 1238.9 1.5103      

1.6928 1.8429 1365.2 1.6533 1.4584 1360 1.6253      

1.7758 2.2028 1471.7 1.7381 1.7526 1468.6 1.7117 1.3023 1463.2 1.577

1.8132 2.3774 1525 1.7758 1.8942 1522.4 1.7499 1.411 1518 1.7159

Chapter 7—Self-Assessment Problems and Questions   279

h400 psia and 1000°F = ((1525 BTU/lbm – 1522.4 BTU/lbm)/(450 psia - 360 psia)) (450 psia - 400 psia) + 1522.4 BTU/lbm = 1523.84 BTU/lbm The final step in the double interpolation process, as applied in this case, involves interpolating between h400 psia and 1000°F and h400 psia and 900°F, the enthalpy values derived in the first two steps above, to obtain the desired final enthalpy h400 psia and 950°F. The formula for this final step is as follows: h400 psia and 950°F = ((h400 psia and 1000°F   - h400 psia and 900°F)/( 1000°F - 900°F))( 950°F - 900°F)   + h400 psia and 900°F Substituting enthalpy values derived in the first two steps above: h400 psia and 950°F = ((1523.84 BTU/lbm   - 1470.32 BTU/lbm)/(1000°F - 900°F))( 950°F - 900°F)   + 1470.32 BTU/lbm = 1497.08 BTU/lbm 3.

In Case Study 7.1, as an energy engineer you have been retained by Station Zebra to explore or develop an alternative integrated steam turbine and electric power generating system that is capable of generating 10 MW of power with only 60 truckloads, or 54,432 kg, of ice per hour. With all other parameters the same as in the original Case Study 7.1 scenario, determine the total heat flow rate, in kJ/hr, needed to produce 10 MW of electrical power.

Solution This problem requires accounting for heat added during each of the five (5) stages of the overall process, with a working fluid mass flow rate of 60 truckloads, or 54,432 kg, per hour. Therefore, the solution is divided into five subparts, each involving either a sensible or a latent heat calculation, based on the entry and exit temperature and phase status. Table 7.2 lists specific heat for water and ice. These heat values will be used in the sensible heat calculations. Table 7.3 lists latent heat values for

280  Appendix A

Figure A-7.4  Case study 7.1 Stage 1 sensible heat calculation.

water. These values will be used to compute the latent heats associated with stages that involve phase transformation. (i)  Calculate the heat required to heat the ice from -10°C to 0°C. Since there is no change in phase involved, the entire heat absorbed by the ice (working substance) in this stage would be sensible heat. First stage of the overall power generating system is illustrated in Figure A-7.4, below. Given: Ti = -10°C Tf = 0°C cice = 2.1 kJ/kg °K 

{Table 7.3}

Utilizing the given information: ΔT = Tf - Ti \ ΔT = 0 – (-10°C) = +10°C Since ΔT represents the change in temperature and not a specific absolute temperature, \ ΔT = +10°K Mathematical relationship between sensible heat, mass of the working substance, specific heat of the working substance and change in temperature can be stated as: Qs(heat ice) = mciceΔT(7.14)

Chapter 7—Self-Assessment Problems and Questions   281

And, Q̇ s(heat ice) = ṁciceΔT 

(7.15)

Where, Qs(heat ice) = Sensible heat required to heat the ice over ΔT Q̇ s(heat ice) = Sensible heat flow rate required to heat the ice over ΔT m = Mass of ice being heated cice = Specific heat of ice = 2.1 kJ/kg °K = Change in temperature, in °C or °K ΔT ṁ = Mass flow rate of water/ice = 60 tons/hr = (60 tons/hr)(907.2 kg/ton) = 54,432 kg/hr Then, by application of Eq. 7.15: Q̇ s(heat ice) = (54,432 kg/hr)(2.1 kJ/kg °K)(10°K) Or, Q̇ s(heat ice) = 1,143,072 kJ/hr Since there are 1.055 kJ per BTU, Q̇ s(heat ice) = (1,143,072 kJ/hr)/(1.055 kJ/BTU) Or, Q̇ s(heat ice) = 1,083,481 BTU/hr (ii)  Calculate the heat required to melt the ice at 0°C. Since change in phase is involved in this case, the heat absorbed by the ice (working substance) in this stage would be latent heat. The 2nd stage of the overall power generating system is illustrated in Figure A-7.5, below.

282  Appendix A

Figure A-7.5  Case study 7.1 Stage 2 latent heat calculation.

Mathematical relationship between latent heat, mass of the working substance, and the heat of fusion of ice can be stated as: Ql(latent ice) = hsl (ice) m 

(7.16)

Q̇ l(latent ice) = hsl (ice)ṁ 

(7.17)

And,

Where, Ql(latent ice) = Latent heat required to melt a specific mass of ice, isothermally Q̇ l(latent ice) = Latent heat flow rate required to melt a specific mass of ice, isothermally, over a period of time m = Mass of ice being melted ṁ = Mass flow rate of water/ice = 60 tons/hr = (60 tons/hr)(907.2 kg/ton) = 54,432 kg/hr, same as part (a) (i) hsl (ice) = Heat of fusion for Ice = 333.5 kJ/kg {Table 7.3} Then, by application of Eq. 7.17: =h ṁ Q̇ l(latent ice)

sl (ice)

Q̇ l(latent ice) = (333.5 kJ/kg)(54,432 kg/hr) = 18,153,072 kJ/hr Q̇ l(latent ice)

Chapter 7—Self-Assessment Problems and Questions   283

Figure A-7.6  Case study 7.1 Stage 3 sensible heat calculation.

Since there are 1.055 kJ per BTU, = (18,153,072 kJ/hr)/(1.055 kJ/BTU) Q̇ l(latent ice)

Or, Q̇ l(latent ice) = 17,206,703 BTU/hr Note that the specific heat required to melt ice is called heat of fusion because of the fact that the water molecules come closer together as heat is added in the melting process. The water molecules are held apart at specific distances in the crystallographic structure of solid ice. The heat of fusion allows the molecules to overcome the crystallographic forces and “fuse” to form liquid water. This also explains why the density of water is higher than the density of ice. (iii)  Calculate the heat reqd. to heat the water from 0°C to 100°C The 3rd stage of the overall power generating system is illustrated in FigureA-7.6, below. Since no phase change is involved in this stage, the heat absorbed by the water in this stage would be sensible heat. Given: = 0°C Ti = 100°C Tf cp-water = 4.19 kJ/kg °K  Utilizing the given information: ΔT = Tf - Ti \ ΔT = 100°C – 0°C = 100°C

{Table 7.3}

284  Appendix A Since ΔT represents the change in temperature and not a specific absolute temperature, \ ΔT = 100°K Mathematical relationship between sensible heat, mass of the working substance, specific heat of water (the working substance), and change in temperature can be stated as: Qs(water) = mcp-water ΔT(7.18) And, Q̇ s(water) = ṁcp-water ΔT 

(7.19)

Where, Qs(water) = Sensible heat required to heat the water over ΔT Q̇ s(water) = Sensible heat flow rate required to heat the water over ΔT m = Mass of water being heated cp-water = Specific heat of water = 4.19 kJ/kg °K ṁ = Mass flow rate of water = 54,432 kg/hr, as calculated in part (a) ΔT = Change in temperature, in °C or °K Then, by applying Eq. 7.19: = ṁc ΔT Q̇ s(water)

p-water

Q̇ s(water) = (54,432 kg/hr)(4.19 kJ/kg °K)(100°K) = 22,807,008 kJ/hr Q̇ s(water)

Since there are 1.055 kJ per BTU, = (22,807,008 kJ/hr)/(1.055 kJ/BTU) Q̇ s(water)

Or, Q̇ s(water) = 21,618,017 BTU/hr iv) Calculate the heat required to convert 100°C water to 100°C steam The 4th stage of the overall power generating system is illustrated in Figure A-7.7, below. Since change in phase is involved in this case, the heat absorbed by the water in this stage would be latent heat.

Chapter 7—Self-Assessment Problems and Questions   285

Figure A-7.7  Case study 7.1 Stage 4 latent heat calculation.

Mathematical relationship between latent heat of vaporization for water, hfg(water), mass of the water, and the total heat of vaporization of water, Ql(latent water), can be stated as: Ql(latent water) = hfg (water) m 

(7.20)

Q̇ l(latent water) = hfg (water) ṁ 

(7.21)

And,

Where, Q l(latent water) = Latent heat of vaporization of water required to evaporate a specific mass of water, isothermally Q̇ l(latent water) = Latent heat of vaporization flow rate required to evaporate a specific mass of water, isothermally, over a given period of time m = Mass of water being evaporated ṁ = Mass flow rate of water = 60 tons/hr = (60 tons/hr)(907.2 kg/ton) = 54,432 kg/hr, same as part (a) (i) hfg (water) = latent heat of vaporization for water = 2257 kJ/kg  {From the steam tables and Table 7.3} Then, by application of Eq. 7.21: =h ṁ Q̇ l(latent water)

fg (water)

Q̇ l(latent water) = (2257 kJ/kg)(54,432 kg/hr) = 122,853,024 kJ/hr Q̇ l(latent water)

286  Appendix A

Figure A-7.8  Case study 7.1 Stage 5 sensible heat calculation.

Since there are 1.055 kJ per BTU, Q̇ l(latent water) = (122,853,024 kJ/hr)/(1.055 kJ/BTU) Or, Q̇ l(latent water) = 116,448,364 BTU/hr (v)  Calculate the heat reqd. to heat the steam from 100°C, 1-atm (102 KPa, or 1-bar) to 500°C, 2.5 MPa superheated steam The 5th stage of the overall power generating system is illustrated in Figure A-7.8, below. Since this stage involves no phase change, the heat absorbed by the steam is sensible heat. In superheated steam phase, the heat required to raise the temperature and pressure of the steam can be determined using the enthalpy difference between the initial and final conditions. Given: Ti = 100°C Pi = 1 - Atm. Note: At 100°C, the saturation pressure is 1- Atm, 1-Bar, or 102 kPa Tf = 500°C Pf = 2.5 MPa For the initial and final temperature and pressure conditions stated above, the enthalpy values, as read from saturated steam table excerpt in Table 7.4a and the superheated steam table excerpt in Table 7.4, are as follows: hi = 2676 kJ/kg at 100°C, 1-Atm hf = 3462 kJ/kg at 500°C, 2.5 MPa

Table 7.4a  Excerpt, saturated steam table, SI units.

Specific Volume m3/kg

Enthalpy kJ/kg

Entropy kJ/kg.°K

Temp. °C

Abs. Press. MPa

20

0.002339

0.0010018

57.7610

83.920

2454.1

2537.5

0.2965

8.6661

20

50

0.012351

0.0010121

12.0280

209.34

2382.7

2591.3

0.7038

8.0746

50

100

0.101420

0.0010435

1.6719

419.10

2257.0

2675.6

1.3070

7.3541

100

200

1.554700

0.0011565

0.1272

852.39

1940.7

2792.1

2.3308

6.4303

200

Sat. Liquid Sat. Vapor Sat. Liquid nL nV hL

Evap. hfg

Sat. Vapor Sat. Liquid Sat. Vapor Temp. hV SL SV °C

Chapter 7—Self-Assessment Problems and Questions   287

Properties of Saturated Steam by Temperature Metric/SI Units

288  Appendix A Equations for determining the heat required to boost the steam from 100°C, 1-Atm to 500°C, 2.5 MPa are as follows: ΔQ steam = (hf - hi )m 

(7.22)

Q̇ steam= (hf - hi)ṁ(7.23) Where, ΔQ steam = Addition of heat required for a specific change in enthalpy Q̇ steam = Rate of addition of heat for a specific change in enthalpy hi = Initial enthalpy hf = Final enthalpy m = Mass of steam being heated ṁ = Mass flow rate of steam as calculated in part (a) of this case study = 54,432 kg/hr  {From Part (a)} Then, by applying Eq. 7.23: = (h - h )ṁ Q̇ steam

f

i

Q̇ steam = (3462 kJ/kg - 2676 kJ/kg)(54,432 kg/hr) = 42,783,552 kJ/hr Q̇ steam

Since there are 1.055 kJ per BTU, = (42,783,552 kJ/hr)/(1.055 kJ/BTU) Q̇ s(water)

Or, Q̇ s(water) = 40,553,130 BTU/hr After assessing the heat added, per hour, during each of the five (5) stages of the steam generation process, add all of the heat addition rates to compile the total heat addition rate for the power generating station. The tallying of total heat is performed in BTU’s/hr as well as kJ/hr. Total Heat Addition Rate in kJ/hr: Total Heat Required to Generate 500°C, 2.5 MPa steam from -10°C Ice, at 54,432 kg/hr = 1,143,072 kJ/hr + 18,153,072 kJ/hr + 22,807,008 kJ/hr   + 122,853,024 kJ/hr + 42,783,552 kJ/hr = 207,739,728 kJ/hr

Chapter 7—Self-Assessment Problems and Questions   289

Total Heat Addition Rate in BTU’s/hr: Total Heat Required to Generate 500°C, 2.5 MPa steam from -10°C Ice, at 54,432 kg/hr = 1,083,481 BTU/hr +17,206,703 BTU/hr +21,618,017 BTU/hr   +116,448,364 BTU/hr +40,553,130 BTU/hr = 196,909,695 BTU/hr 4.

If all of the working fluid, or steam, discharged from the turbine in Case Study 7.1 is reclaimed, reheated, and returned to the turbine, what would be the overall system efficiency? Assume that the mass flow rate is 58,860 kg/hr, or 65 tons per hour.

Solution Strategy Energy and process flow pertaining to this special case scenario of Case Study 1 is depicted in Figure A-7.9. In order to derive the efficiency for this scenario where all of the steam discharged from the turbine is reclaimed and used as working fluid fed into the boiler in the last, superheating, stage, we need to determine the rate of heat addition required to raise the temperature from 150°C to 500°C and the pressure from 50 kPa to 2.5 MPa. This rate of addition of heat can be determined using the following formula: Q̇ steam= (hf - hi )ṁ 

Figure A-7.9  Case study 7.1, Mass flow rate analysis.

(7.23)

290  Appendix A Once Q̇ steam is determined, we can convert it into equivalent power (kW or MW) units for computation of efficiency through the following formula:

Total Station Energy Efficiency, in Percent = Power Output / Power Input * 100

Or,

Total Station Energy Efficiency, in Percent = (Power Output in MW / Q̇ ) * 100 steam in MW

Note: With the exception of the provisions stipulated in the problem statement above, all of the pertinent given data from Case Study 1 remains the same, as stated below: Solution: Given: Pshaft ṁ hi hf

= (10 MW) / hg = (10 MW) / (0.9) = 11.11 × 106 W = 11.11 × 106 J/s = 58,860 kg/hr = 2780 kJ/kg = 2780x103 J/kg  = 3462 kJ/kg = 3462x103 J/kg 

Then, by applying Eq. 7.23: Q̇ steam = (hf - hi )ṁ = (3462 kJ/kg - 2780 kJ/kg)58,860 kg/hr = 40,142,520 kJ/hr Or, Q̇ steam in kW = (40,142,520 kJ/hr) / (3600 s/hr)

= 11,151 kJ/s

Or, since 1J/sec = 1W, and 1kJ/s = 1kW, = = 11,151 kW Q̇ steam in kW

Or, Q̇ steam in MW = (11,151 kW)/(1000kW/MW) = 11.15 MW

{See Table 7.4} {See Table 7.4}

Chapter 7—Self-Assessment Problems and Questions   291 Table A-7.4  Excerpt, superheated steam table, SI units.

Abs. Press. in MPa (Sat. T, °C ) 0.05 (81.33)     0.1 (99.61)     1.0 (179.89)     2.5 (223.99)     3.0 (233.86)     4.0 (250.36)  

n h s   n h s   n h s   n h s   n h s   n h s

Properties of Superheated Steam Metric/SI Units “h” in Temp. in “n” in kJ/kg °C m3/kg 150 300 500 3.889 5.284 7.134 2780.1 3075.5 3488.7 7.9401 8.5373 9.1546       1.9367 2.6398 3.5656 2776.6 3074.5 3488.1 7.6147 8.2171 8.8361         0.2580 0.3541   3051.7 3479.0   7.1247 7.7640         0.0989 0.13998   3008.8 3462.1   6.6438 7.3234         0.0812 0.1162   2994.3 3457.0   6.5412 7.2356         0.0589 0.0864   2961.7 3445.8   6.3638 7.0919

Then, Total Station Energy Efficiency, in Percent = Power Output / Power Input * 100 = 10 MW/11.15MW = 89.68%

“s” in kJ/kg °K 650                 0.4245 3810.5 8.1557   0.1623 3799.7 7.7056   0.1405 3797.0 7.6373   0.1049 3790.2 7.4989

  800                 0.4944 4156.2 8.5024   0.1896 4148.9 8.0559   0.1642 4147.0 7.9885   0.1229 4142.5 7.8523

292  Appendix A

Chapter 8—Self-Assessment Problems and Questions 1.

An ideal heat engine always includes a boiler with superheating function.

Answer: B. False. Heat engines equipped with superheating function, albeit common, is just one type of heat engine.

••

2.

In the heat engine represented by the enthalpy versus entropy graph in Figure 8.22, the heat is added to working fluid in:

Answer: (iv) Steam generation stage and the Steam superheating stage

••This is evident from Figure A-8.1. Enthalpy is converted to the kinetic mechanical energy of the turbine in the path 4 to 5

3.

In the heat engine represented by the enthalpy versus entropy graph in Figure 8.22, the energy contained in the superheated working fluid is converted into the rotational kinetic energy in:

Answer: (ii) Process transition from point 4 to 5. Figure A-8.1 is a duplicate of Figure 8.22, referred to in the problem statement. As evident from Figure A-8.1 below, path 4 to 5 represents the conversion of enthalpy contained in the superheated steam into mechanical kinetic energy of the turbine.

••

Figure A-8.1  Heat cycle in a rankine engine with superheat, h versus s.

Chapter 8—Self-Assessment Problems and Questions  293

4.

A thermodynamic system consists of a Rankine engine with superheat function. The enthalpy versus entropy graph for this system is shown in Figure 8.33.

The mass flow rate of the system or working fluid is 100 lbm/s. Answer the following questions based on the data provided: a) If the enthalpy of the fluid is approx. 1850 BTU/lbm and the entropy of the system is approximately 1.8900 BTU/(lbm °R), what phase or stage is the system in: Note: Figure 8.33, referred to in the problem statement is duplicated below in form of Figure A-8.2, with pertinent annotations, for solution illustration purposes. Solution/Answer: (i) Work producing turbine stage The point representing an enthalpy of approximately 1850 BTU/lbm and the entropy of approximately 1.8900 BTU/(lbm °R) is labeled as “a” in the Figure

Figure A-8.2  Heat cycle in a rankine engine with superheat, h versus n..

294  Appendix A A-8.2 enthalpy—entropy diagram. This point is clearly on thermodynamic transition path 4 to 5. This path represents the work producing turbine stage. b) If the enthalpy of the subcooled water entering the boiler is 900 BTU/ lbm and the enthalpy of the water at a downstream point in the boiler is 1080 BTU/lbm, what is approximate amount of heat added in MMBTU per hour? Assume that there is no heat loss. Answer: (ii) 1.08 MMBTU/hr Solution Since we are allowed to assume that there is not heat loss, the difference between the enthalpy of the subcooled water entering the boiler is and the enthalpy of the water at a downstream point in the boiler would be equal to the heat energy added on per lbm basis. Also, the mass flow rate, ṁ, of the working fluid is given. ∴ Heat energy delivered to the water, in BTU’s/sec Q = (hf – hi)(ṁ) Where, ṁ = Mass flow rate of the working fluid or water = 100 lbm/s {Given} hi = Enthalpy of steam entering the turbine = 900 BTU/lbm 1860 {Given} hf = Enthalpy of steam discharged = 1080 BTU/lbm {Given} Then, application of Eq. 8.2 would yield: Heat energy delivered to the water, in BTU’s/s = (1080 BTU/lbm – 900 BTU/lbm)(100 lbm/s) = 18,000 BTU/s Heat energy delivered to the water, in BTU’s/hr = (18,000 BTU/s)(60 s/hr.) = 1,080,000 BTU’s/hr Since there are 1,000,000 BTU’s per MMBTU, Heat energy delivered to the water, in MMBTU’s/hr 1, 080, 000 BTU’s / hr 1, 000, 000 BTU’s = 1.080 MMBTU/hr

=

Chapter 9—Self-Assessment Problems and Questions   295

Therefore, the answer is: (ii) 1.08 MMBTU/hr Ancillary to part (b): As apparent upon examination of Figure A-8.2, the rise in enthalpy stated in part (b) - from 900 BTU/lbm to1080 BTU/lbm - occurs while the working fluid remains in subcooled realm. If the final enthalpy, hf, were in the vapor-liquid mixture region, the solution for this part of the problem would have required more steps. This is because the enthalpy of the working fluid at final point would be a sum of the saturated vapor and liquid enthalpies, added in proportion determined by the humidity ratio, ω. c) If the enthalpy of the working fluid is 1440 BTU/lbm and the entropy is 1.8500 BTU/(lbm-°R), what is the phase of the working fluid? Answer: (iii) A mixture of vapor and liquid To address this question, we must locate the point on the graph where h = 1440 BTU/lbm and s = 1.8500 BTU/(lbm °R). This point is labeled as point “c” on the enthalpy versus entropy graph in Figure A-8.2. Point c lies in the region that falls between saturated liquid line and the saturated vapor line. In other words, at point c, the working fluid is in a phase that consists of a mixture of liquid and vapor.

Chapter 9—Self-Assessment Problems and Questions 1.

A nozzle is fed from a superheated steam reservoir. The superheated steam in the reservoir is at 500°C (932°F) and 2.0 MPa (290 psia). The duct or hose connecting the nozzle to the reservoir is short and the frictional head loss in the hose is negligible. Based on these practical assumptions, the velocity of the superheated steam in the hose can be neglected. The steam exits the nozzle at 1.0 bar (14.5 psia) and 95% quality. Determine the exit velocity of the steam at the nozzle in SI units.

Solution: Given: To = 500°C Po = 2.0 MPa v1 = v0 = 0 x = Quality = 95% P2 = 1.0 bar or 14.5 psia

296  Appendix A SI Unit System Apply Eq. 9.4 to calculate the exit velocity of the superheated steam in the SI units: v2 = 2(h0 - h2 )

{SI Unit System} 

(9.4)

From the steam tables in Appendix B, in the SI units: ho = 3468 kJ/kg From Mollier diagram, in Figure A-9.1: h2 = 2550 kJ/kg Note: The enthalpies are converted into J/kg, from kJ/kg because Eq. 9.4 is premised on Joules and not kilo Joules. Then, by applying Eq. 9.4: v2 = (2).(3468 - 2550kJ / kg ).(1000 J / kJ ) v2 = 1355m / s

2.

Solve Problem 1 in US Units. Use Mollier diagram for all enthalpy identification and compare the resulting steam speed with results from computation conducted in SI units, in Problem 1.

Solution Given: To = 932°F Po = 290 psia v1 = v0 = 0 x = Quality = 95% P2 = 14.5 psia US Unit System Apply Eq. 9.5 do calculate the exit velocity of the superheated steam in the US units: v2 = 2 g c J (h0 - h2 )

{US Unit System}

From the Mollier Diagram, in Appendix B: ho = 1490 BTU/lbm h2 = 1100BTU/lbm

(9.5)

Chapter 9—Self-Assessment Problems and Questions   297

Figure A-9.1  Mollier diagram, self-assessment problem 1.

Then, by applying Eq. 9.5:  lbm - ft   ft - lbf v2 = 2  32.2 778 2  BTU lbf - s   v2 = 2324 ft / s

BTU   1299 - 1191 lbm

298  Appendix A Comparison with the SI (Metric) v2 calculation: v2 from SI (Metric) Unit calculation is 1355 m/s. If this Metric speed is converted to US Units, through simple unit conversion, v2 in US units amounts to: 4442 ft/s v2 from US (Imperial) Unit calculation, as computed above is: 4420 ft/s The difference in nozzle speed calculated in the SI realm versus the US realm, is: = ((4442 ft/s /4420 ft/s) – 1)(100%) = 0.49% This verifies that while the US Unit method, utilizing Eq. 9.5, and the SI Unit method, utilizing Eq. 9.4, yields nozzle speed results that are slightly different, the difference is less than a percent. 3.

The SFEE Equation 9.2 can be applied to compute the exit speed of gas, in high speed gas applications, under which of the following conditions?



A. When data is available in US units



B. When data is available in SI units



C. When the reservoir is large enough such that v0 = 0, applies.



D. Both B and C.



E. Both A and B.

Answer: As explained in Chapter 9, SFEE Equation 9.2 can be applied to compute the exit speed of gas, in high speed gas applications, in the SI Unit System, when the reservoir is large enough such that v0 = 0, applies. Therefore, the correct answer is “D.” 4.

Which of the following statements is true about shock waves?



A. Shock waves require superheated steam



B. Shock waves travel parallel to the direction of the flow of gas.



C. Shock waves travel perpendicular to the direction of the flow of gas.



D. Both A and B.

Answer: C. Shock waves travel perpendicular to the direction of the flow of gas.

••

Chapter 10—Self Assessment Problems and Questions  299

Chapter 10—Self Assessment Problems and Questions Problem 1 1.

[Note: Problem restated for reader’s convenience] In an environment that is estimated to contain, approximately, 400 kg of air, the dry-bulb is measured to be 40°C and the wet-bulb is at 27.3°C. Later, the air is cooled to 20°C and, in the process of lowering the dry-bulb temperature, the relative humidity drops to 47%. As an Energy Engineer, you are to perform the following psychrometric analysis on this system using the psychrometric chart in Figure 10.6:



a. Find the initial humidity ratio, ωf.



b. Find the final humidity ratio, ωf.



c. Find the total amount of heat removed.



d. Find the amount of sensible heat removed.



e. Find the amount of latent heat removed.



f. Find the final wet-bulb temperature.



g. Find the initial dew point.



h. Find the final dew point.



i. Find the amount of moisture condensed/removed.

As explained in Chapter 10 case study solutions, we need to begin the analyses associated with psychrometric problems with the identification or location of initial and final psychrometric points on the psychrometric chart. Location of the initial psychrometric point in this problem can be established using the following two parameters associated with the point:

•• ••

Dry-bulb temperature of 40°C Wet-bulb temperature of 27.3°C

This point is shown on the psychrometric chart in Figure A-10.1, as point A. Location of the final psychrometric point can be established using the following two pieces of data associated with this point:

•• ••

Dry bulb temperature of 20°C Relative humidity of 47%.

Relative humidity line, representing an RH of 47% is placed through interpolation between the given 50% and 40% RH lines on the psychrometric chart.

300  Appendix A The final point would be located at the point of intersection of the 47% RH line and the vertical line representing the dry bulb temperature of 20°C. The final point, thus identified, is shown as point B on the psychrometric chart in Figure A-10.1. a) Find the initial humidity ratio, ωi. To determine the initial humidity ratio, draw a horizontal line from the initial point to the vertical humidity ratio scale, on the psychrometric chart as shown in Figure A-10.1. The point of intersection of this horizontal line and the humidity ratio scale represents the humidity ratio ωi for the initial psychrometric point. As read from the psychrometric chart in Figure A-10.1: ωi = 0.018 kg of moisture per kg of dry air b) Find the final humidity ratio, ωf. Similar to part (a), the humidity ratio for the final psychrometric point can be determined by drawing a horizontal line from the final point to the vertical humidity ratio scale on the psychrometric chart in Figure A-10.1. The point of intersection of this horizontal line and the humidity ratio scale represents the humidity ratio, ωf, for the final psychrometric point. As read from the psychrometric chart in Figure A-10.1: ωf = 0.007 kg of moisture per kg of dry air c) Find the total amount of heat removed. The first step is to identify the enthalpies, on the psychrometric chart, at the initial and final points. See Figure A-10.1. At the initial point, the dry-bulb temperature is 40°C, the wet-bulb is 27.3°C, and as shown on the psychrometric chart, hi = 87 kJ/kg of dry air. At the final point, dry-bulb is 20°C, with RH at 47%. The enthalpy at this point, hf = 38 kJ/kg of dry air. ∴ Δh = hf - hi = 38 kJ/kg - 87 kJ/kg = - 49 kJ/kg of dry air. And, ΔQ = (Δh)mair  = (-49 kJ/kg)mair

(10.1)

Chapter 10—Self Assessment Problems and Questions  301

Where, the mass of dry air, mair, needs to be derived through the given combined mass of moisture and air, 400 kg, (Note: This is the mass of moist air at the initial point) and the initial humidity ratio, ωi. Humidity ratio is defined as: ω = mass of moisture (kg) / mass of dry air (kg) And, as determined from the psychrometric chart, earlier, in part (a): ω = 0.018 kg of moisture per kg of dry air, at the initial point ω = m moisture / m dry air Or, ω = (m moist air - m dry air ) /m dry air Through algebraic rearrangement of this equation, we get: (1 + ω) = m moist air / m air Or, m air = mass of dry air = m moist air / (1 + ω) Where the total combined of mass of the moisture and the dry air is given as 400 kg. ∴ m air = 400 kg/ (1 + 0.018) = 393 kg Then, by applying Eq. 2.1: ΔQ

= Total Heat Removed = (Δh)mair = (- 49 kJ/kg)(393 kg)

Or, ΔQ = Total Heat Removed = - 19,257 kJ The negative sign, in the answer above, signifies that the heat is extracted or that it exits the system as the air conditioning process transitions from the initial psychrometric point to the final psychrometric point. d) Find the amount of sensible heat removed. The first step in determining the amount of sensible heat removed is to identify the SHR, Sensible Heat Ratio, from the psychrometric chart. This process involves drawing a straight line between the initial and final points. This line

302  Appendix A is shown as a dashed line between the initial and final points. Then draw a line parallel to this dashed line such that it intersects with the SHR Reference Point and the vertical scale representing the SHR. The point of intersection reads, approximately, 0.43. See Figure A-2.1. Note: For additional discussion on the significance of SHR, refer to Case Study 2.2, part (g) and the self-assessment problem 10.2 (g). SHR, Sensible Heat Ratio, is defined, mathematically, as: SHR = Sensible Heat / Total Heat In this case, SHR = Qs / Qt = 0.43 Or, Qs = Sensible Heat = (0.43)(Qt) And, since Qt = ΔQ = Total Heat Removed = -19,257 kJ, as calculated earlier, Qs = Sensible Heat = (0.43)(-19,257 kJ ) = -8,281 kJ e) Find the amount of latent heat removed. The total heat removed consists of sensible and latent heat components. Or, Qt = Qs + Ql Ql = Latent Heat = Qt - Qs = -19,257 kJ – (-8,281 kJ) ∴ Ql = Latent Heat Removed = -10,976 kJ f) Find the final wet-bulb temperature. As explained in the discussion associated with the psychrometric chart interpretation guide in Figure 2.2, wet bulb is read diagonally from the psychrometric point along the wet bulb temperature scale. The wet bulb lines run parallel to the enthalpy lines on the psychrometric chart. The diagonal line emerging from the final point intersects the wet bulb and dew point scale at approximately 13.5°C as shown on the psychrometric chart in Figure A-2.1. Therefore, the wet bulb temperature at the final point is 13.5°C.

Chapter 10—Self Assessment Problems and Questions  303

g) Find the initial dew point. As shown on the psychrometric chart in Figure A-2.1, follow the horizontal “dew point” line drawn from the initial point to the left, toward the saturation curve. The point of intersection of the saturation line and the dew point line represents the dew point. This point lies at 23.2°C. Therefore, the dew point at the initial point, as read off from Figure A-2.1, is 23.2°C. h) Find the final dew point. As shown on the psychrometric chart in Figure A-2.1, follow the horizontal “dew point” line drawn from the final point to the left, toward the saturation curve. The point of intersection of the saturation line and the dew point line represents the dew point. This point lies at 8.6°C. Therefore, the dew point at the initial point, as read off from Figure A-2.1, is 8.6°C. i) Find the amount of moisture condensed/removed. In order to calculate amount of moisture condensed or removed, we need to find the difference between the humidity ratios for the initial and final points. Humidity ratio can be defined mathematically as: Humidity Ratio, ω = mass of moisture (kg) / mass of dry air (kg) From the psychrometric chart, in Figure A-2.1: ωi = 0.018 kg of moisture per kg of dry air, at the initial point ωf = 0.007 kg of moisture per kg of dry air, at the final point ∴ Δω = Change in the Humidity Ratio = 0.018 - 0.007 = 0.011 kg of moisture per kg of dry air The amount of moisture condensed or removed = (Δω)(Total mass of Dry Air) = (Δω)(m dry air ) Where, m dry air = 393 kg of dry air, as calculated in part (a) ∴ The amount of moisture condensed or removed = (0.011 kg of moist./kg of dry air)(393 kg of dry air) = 4.323 kg

Figure A-10.1  Psychrometric chart.

304  Appendix A

Chapter 10—Self Assessment Problems and Questions  305

Solution – Problem 2 2.

[Note: Problem restated for reader’s convenience] Psychrometric chart for the initial and final conditions in a commercial warehouse is shown in Figure 10.7. Dry Bulb and Wet Bulb data associated with the initial and final points is labeled on the chart. Assess the disposition and performance of the HVAC system in this building as follows:



a. What is the initial Dew Point?



b. What is the final Dew Point?



c. Based on the results of dew point determination in parts (a) and (b), define the type of thermodynamic process this system undergoes in the transition from initial point to the final point.



d. What is the RH, Relative Humidity, at the initial point?



e. What is the RH, Relative Humidity, at the final point?



f. Determine the SHR for the change in conditions from the initial to the final point.



g. Comment on why the SHR for this scenario is significantly higher than the scenario analyzed in Case Study 10.2?

As explained in the solution segment of the case studies we, typically, begin our analyses of psychrometric problems with the identification or location of the initial and final psychrometric points on the psychrometric chart. However, in this problem, the initial and final points are already plotted on the psychrometric chart provided. See Figures 10.7 and A-10.2. a) What is the initial Dew Point? To determine the initial dew point, read the dew point temperature for the initial point on the psychrometric chart in Figure A-10.2 (below) by following the horizontal “dew point” line drawn from the initial point to the saturation curve, to the left. The point of intersection of the saturation line and the dew point line represents the dew point. The dew point at the initial point, as read off from Figure A-10.2, is 64.2°F. b) Find the final dew point. As with the initial dew point determination, to determine the final dew point, read the dew point temperature for the final point on the psychrometric chart in Figure A-10.2 (below) by following the horizontal “dew point” line drawn

306  Appendix A from the final point to the saturation curve, to the left. The point of intersection of the saturation line and the dew point line represents the dew point. The dew point, at the final point, as read off from See Figure A-10.2, is 64.2°F, which is the same as the initial dew point. c) Based on the results of dew point determination in parts (a) and (b), define the type of thermodynamic process this system undergoes in the transition from initial point to the final point. The initial and final dew points are the same. In other words, the change in conditions from the initial point to the final point represent a true horizontal shift with substantial dry bulb change, some wet bulb change, but no dew point change. This type of transition that represents a substantial dry bulb change, but negligible dew point change, signifies a definite sensible heat change. This transition is identical to path O-G described in the Psychrometric Transition Process section of Chapter 10. This case involves sensible heat removal in cooling the temperature of the air from 101°F dry bulb to 87°F dry bulb. d) What is the RH, Relative Humidity, at the initial point? Use the psychrometric chart interpretation guide in Figure 10.3 to locate the relative humidity (RH) lines on the psychrometric chart. The initial point lies directly on the 30% RH line. Therefore, the RH at the initial point is 30%. e) What is the RH, Relative Humidity, at the final point? Use the psychrometric chart interpretation guide in Figure 10.3 to locate the two closest relative humidity (RH) lines on the psychrometric chart. The final point lies between the 40% and the 50% RH lines. Through interpolation, the relative humidity for the final point is estimated to be approximately 47%. f) Determine the SHR for the change in conditions from the initial to the final point. Draw a straight line between the initial and final points as shown in Figure A10.2. This line is shown as a dashed line between the initial and final points. Then draw a line parallel to this dashed line such that it intersects with the SHR Reference Point and the scale, to the right, representing the SHR. The point of intersection reads, approximately, 1.0. g) Comment on why the SHR for this scenario is significantly higher than the scenario analyzed in Case Study 10.2? The SHR of 1.0 for the scenario portrayed in this problem is substantially higher than the SHR of 0.18 for the scenario in Case Study 10.2. The lesson

Figure A-10.2  Psychrometric chart.

Chapter 10—Self Assessment Problems and Questions  307

308  Appendix A to be taken away from this self-assessment problem is that when the dry bulb changes significantly while the dew point is held steady, the sensible heat plays the predominant role. In other words, when the dry bulb changes significantly, at a relatively constant dew point, the sensible heat to total heat ratio approaches 1.0. In this case, the SHR is actually 1.0. An SHR of 1.0 implies that the entire amount of heat removed in this case was sensible heat and no latent heat was involved. The scenario portrayed in Case Study 10.2 was somewhat opposite. In that, the change in dry bulb was minute while the change in dew point was substantial. So, the substantial change in latent heat observed in Case Study 10.2 was indicative of the predominance of latent heat. The SHR of 0.18 in case Study 10.2 implied that only 18% of the heat involved in the process was sensible, while 82% of the heat involved in the process was latent.

Chapter 11—Self Assessment Problems and Questions Problem 1 (a)  Draw the refrigeration cycle on the given diagram: The process involved in the drawing of the refrigeration cycle is described below: C – D: See the pressure-enthalpy diagram in Figure A-11.1. Locate the 290 psia and 60 psia points along the pressure (vertical) axis of the chart, name these points “C” and “D,” respectively. This is the throttling portion of the refrigeration cycle. Note: HFC-134a, at point C is in, high pressure, saturated liquid phase. Since the throttling process is adiabatic, and D h = 0, draw a straight, vertical, line down from C to D. At point D, HFC-134a would be in liquid-vapor mixture phase. D - A: See Figure A-11.1. The next step involves complete transformation of the refrigerant from low pressure liquid to gaseous phase, through absorption of heat. Therefore, D h ≠ 0 and this is a non-adiabatic, isobaric, process. Draw a straight, horizontal, line from D to A. This step is referred to as the evaporator segment of the refrigeration cycle. This is where the system (refrigerant) performs cooling of the environment as its phase undergoes latent transformation from liquid to gaseous phase.

Figure A-11.1  Pressure - enthalpy diagram, HFC-134a.

Chapter 11—Self Assessment Problems and Questions  309

310  Appendix A A - B: The next step involves the transformation of HFC-134a from low pressure (60 psia) gaseous phase to high (290 psia) pressure gaseous phase. See Figure A-11.1. This is the compression segment of the refrigeration cycle. This segment of the refrigeration cycle is an isentropic process and Ds = 0. Therefore, draw a straight line from point A to B, asymptotic to s = 0.22 BTU/lbm °F. B – C: The next step involves the transformation of HFC-134a from high (290 psia) pressure gaseous phase to high pressure saturated liquid phase. See Figure A-11.1. This segment constitutes the condenser segment of the refrigeration cycle. This is an isobaric process and DP = 0. Therefore, draw a straight line from point B to C, along the P = 290 psia line. (b) What is the change in enthalpy during the expansion process? Solution/Answer: This is a captious question. In that, as evident from Figure A-11.1, during the C-D throttling or expansion process, the enthalpy stays constant, at h = 65 BTU/lbm. Therefore, there is no change in enthalpy during the expansion process, or Dh = 0. (c) Determine the percentages of liquid and vapor at the end of the throttling segment of the refrigeration cycle. Solution: Determine the value of “x,” the quality, at point “D,” from the PressureEnthalpy diagram. See Figure A-11.1. This involves interpolation or derivation of a quality curve that would intersect through point D. This derived or interpolated curve would be almost directly in between the × = 0.4 and × = 0.5 curves. Hence, × = 0.45 at point D. \ x = (mvapor) / (mvapor + mliquid) = 0.45 or 45% In other words, mvapor (%) = 45 And since (%mvapor + %mliquid) = 100%, mliquid (%) = 100 – 45 = 55% (d) How much enthalpy is absorbed by the system (refrigerant) in the evaporation (latent) phase? Solution:

Chapter 11—Self Assessment Problems and Questions  311

This involves step D – A in Figure A-11.1 Pressure-Enthalpy diagram: Dh D – A = hA - hD = 110 – 65 = 45 BTU/lbm (e) How much enthalpy is extracted from the system (refrigerant) in the condensation (condenser) phase of the cycle? Solution: This involves step B– C in Figure A-11.1 Pressure-Enthalpy diagram. In this segment of the refrigeration cycle: Dh B – C = hC - hB = 65 – 122 = -57 BTU/lbm Note that Dh B – C is negative because heat is lost or extracted from the system in this segment of the refrigeration cycle. (f) Determine the percentages of liquid and vapor at B. Solution/Answer: As apparent from Figure A-11.1, Point B lies directly on the saturated vapor line. And, × = 1at all point on the saturated vapor line. According to Eq. 11.1, x = (mvapor) / (mvapor + mliquid) Therefore, at × = 1 (mvapor) / (mvapor + mliquid) = 1 Or, (mvapor) = (mvapor + mliquid) This can only hold true if mliquid = 0%. And if mliquid = 0%, then mvapor = 100%. Another way to interpret and dissect x = 1, would be through revisiting the definition of quality factor x. Since × represents the ratio of the mass of vapor to the total combined mass of vapor and liquid, x = 100% implies that vapor constitutes the combined mass in entirety. Therefore, the answer is: mliquid = 0% and mvapor = 100% (g) Assume that the mass flow rate of refrigerant being cycled in this air-conditioning system is 10 lbm/min and the compressor efficiency is

312  Appendix A 70%. Determine the amount of electrical power demanded by the compressor motor if the compressor motor efficiency is 90%. Solution: The amount of electrical power demanded by the compression cycle: According to Eq. 11.2, net power required to compress the refrigerant: P = Ẇ = ṁ(hB-hA) Where, point B is equivalent to point 4 and point A is equivalent to point 3. And, Electrical Power Demanded from Utility = P ÷ (Motorefficiency) ÷ (Comp.efficiency) ṁ = 10 lbm/min {Given} = (10 lbm/min)/(60 s/min) = 0.167 lbm/s P = (ṁ)(hB - hA) = (0.167 lbm/s)(122 – 110)(BTU/lbm) = 2 BTU/s Since 1055 J = 1 BTU, P = (2 BTU/s)(1055 J/BTU) = 2110 J/s = 2110 W or 2.11 kW ∴ Electrical Power Demanded from Utility = (2.11 kW)(1/0.7)(1/0.9) = 4.49 kW (h) Which leg of the refrigeration cycle would be considered isentropic. Answer: Path A-B of the HFC-134a refrigeration cycle, as shown in Figure A-11.1, adheres to and stays coincident with the entropy line at s = 0.22 BTU/ lbm °F. The entropy stays constant and equal to 0.22 BTU/lbm °F along path A-B. Therefore, path A-B represents the isentropic process in the given refrigeration cycle. Correct answer: I

Chapter 11—Self Assessment Problems and Questions  313

Solution – Problem 2 Provide the mathematical proof for equivalence of 1 tonne (SI/Metric) of refrigeration capacity to 3.86 kW of power: The amount of heat that must be extracted from 0°C water to freeze it to 0°C ice is equal to the amount of heat that must be added to 0°C ice to melt it to 32°F (0°C) water. This heat of fusion for water, in the metric units is 333.5 kJ/kg Therefore, Rate of refrigeration for one tonne of 0°C water = (333.5 kJ/kg)(1000 kg)/24 hr = 3.86 kJ/s = 3.86 k.(J/s) And, since 1 J/sec = 1 Watt, Rate of refrigeration for one tonne of 0°C water = 3.86kW Solution - Problem 3 Given: SEER Rating: 12 BTU/W-h Air Conditioning System Rating: 20,000 BTU/hr Total, Annual, Seasonal Operating Period: 200 days, 10 hours per day. Average, Combined, Electrical Energy Cost Rate: $0.20/kWh Annual Cost of Energy = ($0.20/kWh).(Total Energy Drawn from The Utility, Annually) Total Power Demanded from The Utility = (Air Conditioning System Rating, in BTU/hr)/ (SEER Rating, in BTU /W-hr

(11.9)

Note: Both BTU values in Eq. 11.9 are outputs, while the W-hr value represents the input energy drawn from the line (utility) side of the power distribution system. \Total Power Demanded from The Utility = (20,000 BTU/hr)/(12 BTU /W hr) = 1,667 Watts, or 1.67 kW

314  Appendix A Then, Total Energy Drawn from The Utility, Annually = (1.667 kW)(Total Annual Operating Hours) = (1.667 kW)(200 Days).(10 Hours/Day) = 3,334 kWh \Total Annual Cost of Electrical Energy Consumed = (3,334 kWh)/($0.20/kWh) = $666.67

Appendix B

Steam Tables These steam tables, copyright ASME, published with ASME permission, do not include the heat of evaporation value, hfg, values for the saturation temperature and pressures. The saturated steam tables presented in this text are the compact version. However, the hfg values for the saturation temperatures and pressures in these tables can be derived by simply adding the available values of hL and hv for the respective saturation pressures and temperatures. In other words: hfg = hL + hv

315

316  Appendix B Table 1. Properties of Saturated Water and Steam (Temperature) Temp. "F

Pressure psi a

Volume, ft3/1bm

Enthalpy, Btu/ lbm

VL

Vy

hl

hv

st.

Sv

Temp. "F

Entropy, Btu/( 1bm· 0 R)

32 35 40 45 50

0.08865 0.09998 0.12173 0.14757 0.17813

0.016022 0.016020 0.016020 0.016021 0.016024

3302.0 2945.5 2443.4 2035.6 1702.9

-0018 3004 8032 13052 18066

1075.2 1076.5 1078.7 1080.9 1083.1

0.0000 0.0061 0.0162 0.0262 0.0361

2.1868 2.1762 2.1590 2.1421 2.1257

32 35 40 45 50

55 60 65 70 75

0.21414 0.25639 0.30579 0.36334 0.43015

0.016029 1430.3 0.016035 1206.1 0.016043 10208 0.016052 867.19 0.016062 739.30

23.074 28079 33080 38078 43074

1085.3 1087.4 1089.6 1091 .8 1094.0

0.0459 0.0555 0.0651 0.0746 0.0840

2.1097 20941 20788 20640 2.0495

55 60 65 70 75

80 85 90 95 100

0.50744 0.59656 0.69899 081636 0.95044

0.016074 0.016086 0.016100 0.016115 0.0161 31

632.44 542.84 467.45 403.79 34987

48069 53062 58054 63.046 68037

1096.1 1098.3 1100.4 1102.6 1104.7

0.0933 0.1025 0.1116 0.1207 0.1296

20353 20215 20080 1.9948 1.9819

80 85 90 95 100

105 110 115 120 125

1.1032 1.2766 1.4730 1.6949 1.9449

0.016148 0.016166 0.016185 0.016205 0.016225

304.05 264.99 231.60 202.96 178.34

73.028 78.019 83010 88002 92.994

1106.9 1109.0 1111 .1 1113.2 1115.3

0.1385 0.1473 0.1560 0.1647 0.1732

1.9693 1.9570 1.9450 1.9333 1.9218

105 110 115 120 125

130 135 140 145 150

2.2258 2.5407 2.8929 3.2858 3.7231

0.016247 0.016269 0.016293 0.016317 0.016342

157.10 138.74 122.82 108.99 96.934

97.987 102.98 107.98 112.97 117.97

1117.4 1119.5 11 21.6 11 23.7 1125.7

0.1817 0.1902 0.1985 0.2068 0.2151

1.9106 1.8996 1.8888 1.8783 1.8680

130 135 140 145 150

155 160 165 170 175

4.2089 4.7472 5.3426 5.9998 6.7237

0.016367 0.016394 0.016421 0.016449 0.016478

86.405 77.186 69097 61.982 55.710

122.97 127.98 132.98 137.99 143.00

1127.8 1129.8 1131.9 1133.9 1135.9

0.2232 0.2313 0.2394 0.2474 0.2553

1.8580 1.8481 1.8384 1.8290 1.8197

155 160 165 170 175

180 185 190 195 200

7.5196 8.3930 9.3497 10.396 11.538

0.016507 0.016538 0.016569 0.016601 0.016633

50.171 45.267 40.918 37.053 33.611

14801 153.03 15805 16307 168.10

1137.9 1139.9 1141 .8 1143.8 1145.7

0.2631 0.2709 0.2787 0.2864 0.2940

1.8106 1.8017 1.7930 1.7844 1.7760

180 185 190 195 200

205 210 215 220 225

12.782 14.136 15.606 17.201 18.928

0.016667 0.016701 0.016736 0.016771 0.016808

30.540 27.796 25.339 23.135 21.155

173.13 178.17 183.20 188.25 193.30

1147.6 1149.5 1151.4 1153.3 1155.1

0.3016 0.3092 0.3167 0.3241 0.3315

1.7678 1.7597 1.7517 1.7440 1.7363

205 210 215 220 225

230 235 240 245 250

20.795 22.811 24.985 27.326 29.843

0.016845 0.016883 0.016921 0.016961 0.017001

19.373 17.766 16.316 15 004 13.816

198.35 203.41 208.47 213.54 218.62

1157.0 1158.8 1160.5 1162.3 1164.0

0.3388 0.3461 0.3534 0.3606 0.3678

1.7288 1.7214 1.7141 1.7070 1.7000

230 235 240 245 250

Steam Tables  317 Table 1. Properties of Saturated Water and Steam (Temperature) Temp.

Of

Pressure psi a

255 260 265 270 275

Volume, ft3jJbm

Enthalpy, Btujlbm

Entropy, Btuj(lbm· 0 R)

Temp.

VL

Vv

hl

hv

SL

Sv

Of

32.546 35.445 38.551 41 .874 45.426

0017042 0.017084 0.017127 0.017170 0.017214

12.739 11 .760 10.870 10.059 9.3196

223.70 228.79 233.88 238.99 244.1 0

1165.7 1167.4 1169.1 1170.7 1172.3

0.3749 0.3820 0.3890 0.3960 0.4030

1.6930 1.6862 1.6796 1.6730 1.6665

255 260 265 270 275

280 285 290 295 300

49.218 53.261 57.567 62.150 67.021

0.017259 0.017305 0.017352 0.017400 0.017449

8.6442 8.0265 7.4610 6.9425 6.4666

249.21 254.34 259.47 264.61 269.76

1173.9 1175.5 1177.0 1178.5 1180.0

0.4099 0.4168 0.4236 0.4305 0.4372

1.6601 1.6538 1.6476 1.6414 1.6354

280 285 290 295 300

305 310 315 320 325

72.193 77.680 83.496 89.654 96.168

0.017498 0.017548 0.017600 0.017652 0.017705

60293 5.6270 5.2564 4.9148 4.5994

274.91 280.08 285.26 290.44 295.64

11814 1182.8 1184.2 1185.5 1186.8

0.4440 0.4507 0.4574 0.4640 0.4706

1.6294 1.6235 1.6177 1.6120 1.6063

305 310 315 320 325

330 335 340 345 350

103.05 110.32 118.00 126.08 134.60

0.017760 0.017815 0.017871 0.017929 0.017987

4.3079 4.0384 3.7888 3.5574 3.3428

300.85 306.07 31 1.30 316.54 321.79

1188.0 1189.3 1190.5 1191.6 1192.7

0.4772 0.4838 0.4903 0.4968 0.5033

1.6007 1.5952 1.5897 1.5843 1.5789

330 335 340 345 350

355 360 365 370 375

143.57 153.00 162.92 173.33 184.25

0.018047 0.018108 0.018170 0.018233 0018297

3.1435 2.9582 2.7859 2.6254 2.4758

32706 332.34 337.63 342.94 348.26

1193.8 1194.8 1195.8 1196.7 11976

0.5097 0.5162 0.5226 0.5289 0.5353

1.5736 1.5684 1.5632 1.5580 1.5529

355 360 365 370 375

380 385 390 395 400

195.71 207.72 220.29 233.45 247.22

0.018363 0.018430 0.018498 0.018568 0.018639

2.3363 2.2061 2.0843 19705 1.8640

353.59 358.94 364.31 369.70 375.10

1198.5 1199.3 1200.1 1200.8 1201 .5

0.5416 0.5479 0.5542 0.5605 0.5667

1.5478 1.5428 1.5378 1.5329 1.5280

380 385 390 395 400

405 410 415 420 425

261.61 276.64 292.34 308.71 325.79

0.018711 0.018785 0.018861 0.018938 0.019016

1.7643 1.6708 1.5830 1.5007 1.4234

380.52 385.95 391.41 396.89 402.38

1202.1 1202.6 1203.2 1203.6 1204.0

0.5729 0.5791 0.5853 0.5915 0.5977

1.5231 1.5182 1.5134 1.5086 1.5038

405 410 415 420 425

430 435 440 445 450

343.59 362.13 381.44 401.53 422.42

0.019097 0.019179 0.019263 0.019349 0.019437

1.3507 1.2822 12179 1.1572 1.1000

40790 413.44 419.01 424.59 430.20

1204.4 1204.7 1204.9 1205.1 1205.2

0.6038 0.6100 0.6161 0.6222 0.6283

1.4991 1.4943 1.4896 1.4849 1.4802

430 435 440 445 450

455 460 465 470 475

444.14 466.71 490.15 514.48 539.73

0.019527 0.019619 0.019713 0.019810 0.019908

1.0461 0.9952 0.9471 0.9016 08586

435.84 44150 447.19 452.91 458.66

1205.2 1205.2 1205.1 1204.9 1204.7

0.6344 0.6405 0.6466 0.6526 0.6587

1.4755 1.4709 1.4662 1.4615 1.4569

455 460 465 470 475

318  Appendix B Table 1. Properties of Saturated Water and St eam (Temperature) Temp.

Of

Pressure psia

480 485 490 495 500

Volu me, ft3j lbm

Ent halpy, Btuj lbm

Entropy, Bt u/( lbm· 0 R) Te mp. Of Sv ~

VL

Vv

hl

hv

565.92 593.07 621.20 650.35 680.53

0.02001 0.02011 0.02022 0.02033 0.02044

0.8180 0.7795 0.7430 0.7084 0.6756

464.44 470.25 476.10 481.97 487.89

1204.4 1204.0 1203.5 1203.0 1202.3

0.6648 0.6708 0.6769 0.6829 0.6890

1.4522 1.4475 1.4429 1.4382 1.4335

480 485 490 495 500

505 510 515 520 525

711.77 744.09 777.52 812.08 847.81

0.02056 0.02068 0.02080 0.02092 0.02105

0.6445 0.6149 0.5868 0.5601 0.5347

493.84 499.83 505.86 511.93 518.05

12016 1200.8 1199.9 1198.9 1197.9

0.6951 0.701 1 0.7072 0.7133 0.7194

1.4288 1.4241 1.4193 1.4145 1.4098

505 510 515 520 525

530 535 540 545 550

884.73 922.85 962.23 1002.9 1044.8

0.021 18 0.02132 0.02146 0.02161 0.02176

0.5105 0.4875 0.4656 0.4446 0.4247

524.21 530.42 536.69 543.00 549.37

1196.7 1195.4 1194.0 1192.5 1190.8

0.7255 0.7316 0.7377 0.7438 0.7500

1.4049 1.4001 1.3952 1.3903 1.3853

530 535 540 545 550

555 560 565 570 575

1088.1 1132.7 11 78.7 1226.2 1275.1

0.02192 0.02208 0.02225 0.02242 0.02260

0.4056 0.3875 0.3701 0.3535 0.3376

555.80 562.29 568.85 575.48 582.18

1189.1 1187.2 1185.2 1183.0 1180.7

0.7562 0.7624 0.7686 0.7749 0.7812

1.3803 1.3752 1.3701 1.3649 1.3596

555 560 565 570 575

580 585 590 595 600

1325.4 1377.3 1430.8 1485.8 1542.5

0.02279 0.02299 0.02319 0.02341 0.02363

0.3223 0.3077 0.2938 0.2803 0.2675

588.95 595.81 602.75 609.79 616.93

1178.2 1175.6 1172.8 1169.8 1166.6

0.7875 0.7939 0.8003 0.8067 0.8133

1.3543 1.3489 1.3433 1.3377 1.3320

580 585 590 595 600

605 610 615 620 625

1600.8 1660.8 1722.6 1786.1 1851.5

0.02387 0.02411 0.02437 0.02465 0.02494

0.2551 0.2432 0.2317 0.2207 0.2101

624.17 631.53 639.01 646.62 654.38

1163.2 1159.6 1155.7 11516 1147.2

0.8198 0.8265 0.8332 0.8400 0.8469

1.3261 1.3202 1.3140 1.3077 1.3012

605 610 615 620 625

630 635 640 645 650

1918.8 1988.0 2059.2 2132.4 2207.7

0.02525 0.02558 0.02593 0.02631 0.02672

0.1998 0.1899 0.1802 0.1709 0.1618

662.30 670.40 678.69 687.21 695.99

1142.5 1137.4 1132.0 1126.1 1119.7

0.8539 0.8610 0.8683 0.8757 0.8833

1.2945 1.2876 1.2804 1.2729 1.2651

630 635 640 645 650

655 660 665 670 675

2285.2 2364.8 2446.8 2531.2 2618.0

0.02717 0.02766 0.02821 0.02883 0.02953

0.1530 0.1444 0.1359 0.1276 0.1194

705.06 714.47 724.30 734.63 745.57

111 2.8 1105.3 1097.2 1088.1 1078.0

0.891 1 0.8991 0.9075 0.9163 0.9255

1.2569 1.2482 1.2390 1.2292 12185

655 660 665 670 675

680 685 690 695 700

2707.3 2799.3 2894.2 2991.9 3092.9

0.03035 0.03133 0.03256 0.03422 0.03683

0.1112 0.1030 0.09444 0.08531 0.07466

757.30 770.10 784.45 801.35 823.64

1066.6 1053.5 1037.9 1018.3 990.64

0.9354 0.9462 0.9582 0.9723 0.9910

1.2068 1.1937 1.1786 1.1602 1.1350

680 685 690 695 700

705

3197.9 3200.1

0.04662 0.0497

0.05338 0.0497

882.44 897.48

913.89 897.48

1.0409 1.0538

1.0679 1.0538

705 T.

Tc

Tc=705.1028 °F

Steam Tables  319 Table 2. Properties of Saturated Water and Steam (Pressure) Pressure psia

Temp.

OF

Volume, ft3jlbm VL

0.1 0.2 0.3 0.5 0.7

35.00 53.13 64.45 79.55 90.05

1.0 1.5 2.0 3.0 4.0

101.69 115.64 126.03 14142 152.91

0.016137 0.016187 0.016230 0.016299 0.016356

6 8 10 12 14

17000 182.81 193.16 201.91 209.52

16 18 20 25 30

Vv

Enthalpy, Btujlbm

hl

hv

Entropy, Btuj(lbm· 0 R) Pressure psia SL Sv

3.009 21.204 32.532 47.618 58.100

1076.5 1084.4 1089.4 1095.9 1100.4

0.0061 0.0422 0.0641 0.0925 0.1117

2.1762 2.1156 2.0805 2.0366 2.0079

0.1 0.2 0.3 0.5 0.7

333.51 227.68 173.72 118.70 90.628

69.728 83.650 94.019 109.39 120.89

1105.4 1111.4 1115.8 1122.2 1126.9

0.1326 0.1571 0.1750 0.2009 0.2198

1.9776 1.9435 1.9195 1.8858 1.8621

1.0 1.5 2.0 3.0 4.0

0.016449 0.016524 0.016589 0.016646 0.016697

61.979 47.345 38.423 32.398 28.048

137.99 150.83 161.22 170.02 177.68

1133.9 1139.0 1143.1 1146.4 1149.4

0.2474 0.2675 0.2836 0.2969 0.3084

1.8290 1.8056 1.7875 1.7728 1.7605

6 8 10 12 14

21627 222.36 227.92 240.03 250.30

0.016745 0.016788 0.016829 0.016922 0.017003

24.755 22.173 20.092 16.306 13.748

184.49 190.63 196.25 208.51 218.93

1151.9 1154.2 1156.2 1160.5 1164.1

0.3186 0.3276 0.3358 0.3534 0.3682

1.7497 1.7403 1.7319 1.7141 1.6995

16 18 20 25 30

35 40 45 50 55

259.25 267.22 274.42 280.99 287.06

0.017078 0.017146 0017209 0.017268 0.017325

11.900 10.500 9.4023 8.5171 7.7878

228.03 236.15 243.50 250.23 256.45

1167.2 1169.8 1172.2 1174.2 1176.1

0.3809 0.3921 0.4022 0.4113 0.4196

1.6873 1.6766 1.6672 1.6588 1.6512

35 40 45 50 55

60 65 70 75 80

292.69 297.96 302.92 307.59 312.03

0017378 0.017429 0.017477 0.017524 0.017569

7.1762 6.6557 6.2071 5.8164 5.4730

262.24 267.66 272.76 277.59 282.18

1177.8 1179.4 1180.8 1182.1 1183.3

0.4273 0.4345 0.4412 0.4475 0.4534

1.6443 1.6378 1.6319 1.6264 1.6212

60 65 70 75 80

85 90 95 100 110

316.25 320.27 324.12 327.82 334.78

0.017613 0.017655 0.017696 0.017736 0.017813

5.1686 4.8969 4.6528 4.4324 4.0496

286.55 290.73 294.73 298.57 305.84

1184.5 1185.6 1186.6 1187.5 1189.2

0.4590 0.4644 0.4695 0.4744 0.4835

1.6163 1.6117 1.6073 1.6032 1.5954

85 90 95 100 110

120 130 140 150 160

341.26 347.33 353.04 358.43 363.55

0.017886 0.017956 0.018023 0.018089 0.018152

3.7286 3.4554 3.2199 3.0148 2.8345

312.62 318.98 324.99 330.68 336.10

1190.7 1192.1 1193.4 1194.5 1195.5

0.4920 0.4998 0.5072 0.5141 0.5207

1.5883 1.5818 1.5757 1.5700 1.5647

120 130 140 150 160

170 180 190 200 210

368.43 373.08 377.54 381.81 385.92

0.018213 0.018272 0.018330 0.018387 0.018442

2.6746 2.5320 2.4038 2.2880 2.1829

341.27 346.21 350.96 355.53 359.94

1196.5 1197.3 1198.1 1198.8 1199.5

0.5269 0.5328 0.5385 0.5439 0.5491

1.5596 1.5549 1.5503 1.5460 1.5419

170 180 190 200 210

0.016020 2945.0 0.016027 1525.9 0.016042 1039.4 0.016073 641.32 0.016100 466.81

320  Appendix B Table 2. Properties of Saturated Water and Steam (Pressure) Pressure psia

Temp.

Of

VL

220 230 240 250 260

389.89 393.71 397.41 400.98 404.45

0.018496 0.018549 0.018601 0.018653 0018703

270 280 290 300 320

407.82 411.09 414.27 417.37 423.33

340 360 380 400 420

Volume , ft3/ 1bm

Enthalpy, Btujlbm

Entropy, Btuj(lbm· 0 R) Pressure psia st_ Sv

hl

hy

20870 1.9992 1.9184 1.8439 17749

364.19 368.30 372.29 376.16 379.92

1200.1 1200.6 1201.1 1201.6 1202.0

0.5541 0.5588 0.5635 0.5679 0.5723

1.5379 1.5342 1.5305 1.5270 1.5236

220 230 240 250 260

0.018753 0018801 0.018849 0.018897 0.018990

1.7108 1.6512 1.5955 1.5434 1.4487

383.58 387.14 390.61 394.00 400.54

1202.4 1202.8 1203.1 1203.4 1203.9

0.5764 0.5805 0.5844 0.5883 0.5956

1.5203 1.5172 15141 15111 1.5054

270 280 290 300 320

429.01 434.43 439.63 444.63 449.43

0.019081 0.019170 0.019257 0.019343 0.019427

1.3647 1.2898 1.2224 1.1616 1.1064

406.81 412.82 418.60 424.18 429.56

1204.3 1204.6 1204.9 1205.0 1205.1

0.6026 0.6093 0.6156 0.6217 0.6276

1.5000 1.4949 1.4900 1.4853 1.4807

340 360 380 400 420

440 460 480 500 520

454.06 458.53 462.86 467.05 471.11

0.019510 0.019592 0.019672 0.019752 0.019831

1.0560 1.0098 0.9673 0.9282 0.8919

434.78 439.84 444.75 449.53 454.19

1205.2 1205.2 1205.1 1205.0 1204.9

0.6333 0.6387 0.6440 0.6490 0.6540

1.4764 1.4722 1.4682 1.4643 1.4605

440 460 480 500 520

540 560 580 600 620

475.05 478.89 482.62 486.25 489.79

0.019909 0.019987 0.02006 0.02014 0.02022

0.8582 0.8268 0.7976 0.7702 0.7445

458.72 463.15 467.48 471.71 475.85

1204.7 1204.4 1204.2 1203.9 1203.5

0.6588 0.6634 0.6679 0.6723 0.6766

1.4568 1.4532 1.4498 1.4464 1.4430

540 560 580 600 620

640 660 680 700 720

493.24 496.62 499.91 503.14 506.29

0.02029 0.02037 0.02044 0.02051 0.02059

0.7203 0.6976 0.6761 0.6559 0.6367

479.91 483.88 487.79 491.62 495.38

1203.2 1202.8 1202.4 1201.9 1201.4

0.6808 0.6849 0.6889 0.6928 0.6966

1.4398 1.4367 1.4336 1.4305 1.4276

640 660 680 700 720

740 760 780 800 820

509.38 512.40 515.36 518.27 521.12

0.02066 0.02073 0.02081 0.02088 0.02095

0.6185 0.6012 0.5848 0.5692 0.5543

499.08 502.72 506.30 509.83 513.30

1200.9 1200.4 1199.9 1199.3 1198.7

0.7004 0.7040 0.7076 0.7112 0.7146

1.4246 1.4218 1.4190 1.4162 1.4135

740 760 780 800 820

840 860 880 900 920

523.92 526.67 529.37 532.02 534.63

0.02102 0.02110 0.02117 0.02124 0.02131

0.5401 0.5265 0.5135 0.5011 0.4892

516.73 520.10 523.43 526.72 529.96

1198.1 1197.5 1196.8 1196.2 1195.5

0.7181 0.7214 0.7247 0.7279 0.7311

1.4108 1.4082 1.4056 1.4030 1.4005

840 860 880 900 920

940 960 980 1000 1050

537.20 539.72 542.21 544.65 550.61

0.02138 0.02146 0.02153 0.02160 0.02178

0.4777 0.4667 0.4562 0.4461 0.4223

533.17 536.34 539.47 542.56 550.15

1194.8 1194.1 1193.3 1192.6 1190.6

0.7343 0.7374 0.7404 0.7434 0.7507

1.3980 1.3955 1.3930 1.3906 1.3847

940 960 980 1000 1050

Vv

Steam Tables  321 Table 2. Properties of Saturated Water and Steam (Pressure) Pressure psia

Temp.

OF

Volume , ft3/ 1bm VL

Vv

Enthalpy, Btu/ lbm

hl

hv

Entropy, Btu/( 1bm· 0 R) Pressure psia Sv St.

1100 1150 1200 1250 1300

556.35 561 .90 567.26 572.46 577.50

0.02196 0.02214 0.02233 0.02251 0.02270

0.4006 0.3808 0.3625 0.3456 0.3299

557.55 564.77 571.84 578.76 585.55

1188.6 11 86.4 11 84.2 11 81.9 1179.5

0.7578 0.7647 0.7714 0.7780 0.7843

1.3789 1.3733 1.3677 1.3623 1.3570

1100 1150 1200 1250 1300

1350 1400 1450 1500 1550

582.39 587.14 591.76 596.27 600.66

0.02288 0.02307 0.02327 0.02346 0.02366

0.3153 0.3017 0.2890 0.2770 0.2658

592.21 598.77 605.23 611.59 617.87

1177.0 1174.4 1171.8 1169.0 1166.2

0.7905 0.7966 0.8025 0.8084 0.8141

1.3517 1.3465 1.3414 1.3363 1.3312

1350 1400 1450 1500 1550

1600 1650 1700 1750 1800

604.93 609.1 1 613.19 617.18 621.07

0.02386 0.02407 0.02428 0.02449 0.02471

0.2553 0.2453 0.2358 0.2269 0.2184

624.07 630.21 636.28 642.30 648.27

11 63.3 1160.3 1157.2 1154.0 1150.7

0.8197 0.8253 0.8307 0.8361 0.8415

1.3262 1.3212 1.3163 1.3113 1.3063

1600 1650 1700 1750 1800

1850 1900 1950 2000 2050

624.89 628.62 632.27 635.85 639.36

0.02493 0.02516 0.02539 0.02563 0.02588

0.2103 0.2026 0.1952 0.1882 0.1814

654.20 66009 665.96 671.80 677.62

1147.3 1143.8 1140.2 1136.5 1132.7

0.8467 0.8519 0.8571 0.8622 0.8673

1.3014 1.2964 1.2914 1.2864 1.2814

1850 1900 1950 2000 2050

2100 2150 2200 2250 2300

642.81 646.18 649.50 652.75 655.94

0.02614 0.02640 0.02668 0.02696 0.02726

0.1750 0.1687 0.1627 0.1569 0.1514

683.44 689.26 695.09 700.93 706.80

1128.7 1124.6 1120.4 1116.0 1111 .5

0.8724 0.8774 0.8825 0.8875 0.8926

1.2763 1.2711 1.2659 1.2606 1.2553

2100 2150 2200 2250 2300

2350 2400 2450 2500 2550

659.08 662.16 665.19 668.17 671.10

0.02757 0.02789 0.02823 0.02859 0.02897

0.1459 0.1407 0.1356 0.1307 0.1258

712.71 718.67 724.69 730.78 736.97

1106.8 11019 1096.8 1091.5 1086.0

0.8976 0.9027 0.9078 0.9130 0.9183

1.2498 1.2443 1.2387 1.2329 1.2269

2350 2400 2450 2500 2550

2600 2650 2700 2750 2800

673.98 676.81 679.60 682.34 685.03

0.02938 0.02981 0.03028 0.03078 0.03134

0.1211 0.1165 0.1119 0.1074 0.1 029

743.27 749.71 756.32 763.13 770.20

1080.2 1074.1 1067.6 1060.7 1053.4

0.9236 0.9290 0.9346 0.9403 0.9462

1.2208 1.2144 1.2078 1.2009 1.1936

2600 2650 2700 2750 2800

2850 2900 2950 3000 3050

687.69 690.30 692.88 695.41 697.90

0.03195 0.03264 0.03344 0.03438 0.03554

0.09843 0.09391 0.08930 0.08453 0.07945

777.59 785.39 793.75 802.90 813.22

1045.5 1036.8 1027.3 101 6.5 1003.8

0.9524 0.9590 0.9660 0.9736 0.9823

1.1859 1.1776 1.1686 1.1585 1.1 469

2850 2900 2950 3000 3050

3100 3150 3200 Pc

700.35 702.75 705.10 705.1 028

0.03708 0.03947 0.04897 0.0497

0.07381 0.06686 0.05052 0.0497

825.57 842.34 893.85 897.48

988.14 966.17 901.07 897.48

0.9926 1.0068 1.0507 1.0538

1.1328 1.11 33 1.0569 1.0538

3100 3150 3200 Pc

Pc = 3200.11 psia

322  Appendix B Table 3. Pressure psi a (Sat. T)

1

V

s V

(162 18) h

s

10

Steam

Temperature-Degrees Fahrenheit 200 250 350 300

(101 .69) h

5

Su~erheated

V

(193.16) h

s

15

V

20

V

(212.99} h

(227.92) h

25

V

30

V

35

V

40

V

45

V

50

V

55

V

60

V

65

V

70

V

(240.03) h

400

450

500

600

700

800

900

1000

1200

392.53 1150.1 2.0510

422.42 11 72.8 2.0842

452.28 482.11 1195.7 121 8.6 2.1152 2.1445

511 .93 1241.8 2.1723

541.74 1265.1 2.1986

571.55 1288.6 2.2238

631.15 1336.2 2.2710

690.74 1384.6 2.3146

750.32 1433.9 2.3554

809.91 1484.1 2.3937

869.48 1535.1 2.4299

988.64 1640.0 2.4973

78.1 55 1148.5 1.8716

84.220 11 71.7 1.9055

90.248 1194.8 1.9370

96.254 1218.0 1.9665

102.25 1241.3 1.9944

108.23 1264.7 2.0209

11 4.21 1288.2 2.0461

126.1 5 1335.9 2.0934

138.09 150.02 1384.4 1433.7 2.1371 2.1779

161.94 1483.9 2.2162

173.87 1535.0 2.2525

197.71 1640.0 2.3198

38.851 1146.4 1.7926

41.942 1170.2 1.8275

44.993 1193.8 1.8595

48.022 1217.2 1.8893

51.036 1240.6 1.9174

54042 57.042 1264.1 1287.8 1.9440 1.9693

63.030 1335.6 2.0167

69.008 1384.2 2.0605

74.980 1433.5 2.1013

80.949 1483.8 2.1397

86.915 98.841 1534.9 1639.9 2.1760 2.2434

27.846 1168.7 1.7811

29.906 1192.7 1.8137

31.943 1216.3 1.8438

33.966 1239.9 1.8721

35.979 1263.6 1.8989

37.986 1287.3 1.9243

41.988 1335.3 1.9718

45.981 1383.9 2.0156

49.968 1433.3 2.0565

53.950 1483.6 2.0949

57.931 1534.7 2.1312

65.886 1639.8 2.1986

20.796 1167.2 1.7477

22.362 1.7808

23.903 1215.5 1.811 3

25.430 1239.3 1.8398

26.947 1263.0 1.8667

28.458 1286.9 1.8922

31.467 1334.9 1.9398

34.467 1383.6 1.9838

37.461 1433.1 2.0247

40.451 1483.4 2.0631

43.438 1534.6 2.0994

49.408 1639.7 2.1669

16.565 1165.6 1.7213

17.835 1190.4 1.7551

19.079 1214.6 1.7859

20.308 1238.6 1.8146

21.528 1262.5 1.8417

22.741 1286.4 1.8673

25.1 55 1334.6 1.9150

27.559 1383.4 1.9590

29.957 1432.9 2.0000

32.352 1483.3 2.0384

34.743 1534.5 2.0748

39.521 1639.6 2.1422

14.816 1189.3 1.7338

15.863 1213.8 1.7650

16.894 1237.9 1.7939

17.915 1261.9 1.8211

18.930 1286.0 1.8468

20.947 1334.3 1.8947

22.954 1383.1 1.9387

24.955 1432.7 1.9797

26.952 1483.1 2.0182

28.946 1534.3 2.0546

32.930 1639.5 2.1 221

12.659 1188.1 1.7156

13.565 1212.9 1.7472

14.455 1237.2 1.7764

15.334 1261.4 1.8037

16.207 1285.5 1.8295

17.941 1333.9 1.8774

19.664 1382.9 1.9216

21 .381 1432.5 1.9626

23.095 1482.9 2.0011

24.806 1534.2 2.0375

28.222 1639.4 2.1050

11.041 11841 11 86.9 1212.0 1.6996 1.7316

12.625 1236.5 1.7610

13.398 1260.8 1.7885

14.1 65 1285.0 1.8144

15.686 1333.6 1.8625

17.197 1382.6 1.9067

18.702 1432.3 1.9478

20.202 1482.7 1.9863

21.700 1534.0 2.0227

24.691 1639.3 2.0903

9.7814 1185.7 1.6854

10.500 12111 1.71 78

11 .202 1235.9 1.7474

11.893 1260.3 1.7750

12.577 1284.6 1.8010

13.933 1333.2 1.8493

15.278 1382.3 1.8935

16.617 1432.1 1.9347

17.952 1482.6 1.9733

19.285 1533.9 2.0097

21.945 1639.2 2.0772

8.7735 1184.5 1.6724

9.4273 1210.2 1.7053

10.063 1235.1 1.7352

10.688 1259.7 1.7629

11.306 1284.1 1.7891

12.530 1332.9 1.8374

13.743 1382.1 1.8818

14.950 1431.9 1.9229

16.153 1482.4 1.9615

17.353 1533.8 1.9980

19.748 1639.1 2.0656

7.9484 1183.2 1.6606

8.5492 1209.3 1.6939

9.1315 1234.4 1.7240

9.7027 1259.1 1.7520

10.267 1283.6 1.7782

11.382 1332.6 1.8267

12.487 1381.8 1.8711

13.585 1431.7 1.9123

14.680 1482.2 1.9509

15.772 1533.6 1.9874

17.951 1639.0 2.0550

7.2604 1181.9 1.6496

7.8173 1208.4 1.6834

8.3549 1233.7 1.7138

8.8813 1258.6 1.7419

9.4004 1283.2 1.7682

10.425 1332.2 1.8168

11.440 1381.5 1.8613

12.448 1431.4 1.9026

13.453 1482.1 1.9413

14.454 1533.5 1.9777

16.453 . 1638.9 2.0454

6.6776 11 80.5 1.6394

7.1978 1207.4 1.6737

7.6978 1233.0 1.7043

8.1862 8.6673 1258.0 1282.7 1.7326 1.7590

9.6160 1331.9 1.8078

10.554 1381.3 1.8523

11.486 1431.2 1.8937

12.414 1481.9 1.9323

13.340 15.1 85 1533.3 1638.8 1.9688 2.0365

6.6666 1206.5 1.6646

7.1344 1232.3 1.6955

7.5904 1257.4 1.7239

8.9223 1331 .5 1.7994

9.7951 1381.0 1.8440

10.662 1431.0 1.8854

11 .524 1481.7 1.9241

12.384 1533.2 1.9606

(250.30) h

(259.25) h

(267.22) h

(274.42) h

(280.99) h

(287.06) h

(292.69) h

(297.96) h

(302.92) h

v= specific volume, ft3Jibm

1191.6~

h =enthalpy, Btu/lbm

S=

8.0389 1282.2 1.7505

entropy, Btu/(1bm·0 R)

14.099 1638.7 20283

Steam Tables  323 Table 3 {cont inued}.

Su~erheated

Pressure psia (Sat. T)

450

Temperature-Degrees Fahrenheit

350

400

500

Steam

550

600

700

80 V 5.8030 6.2186 6.6220 (312.03) h 1204.5 1230.8 1256.2 s 1.6480 1.6795 1.7082

7.0176 7.4081 1281 .3 1306.1 1.7350 1.7602

7.7949 1330.8 1.7842

8.5614 1380.5 1.8289

V 5.1 307 90 (320.27) h 1202.5 s 1.6330

5.5061 5.8686 1229.3 1255.1 1.6651 1.6943

6.2232 6.5724 1280.3 1305.3 1.7213 1.7466

6.9180 7.6019 8.2794 8.9529 1330.1 1380.0 1430.2 14810 1.7707 1.8156 1.8572 1.8960

100 V 4.5923 (327.82) h 1200.4 s 1.6194

4.9358 5.2658 5.5875 1227.8 1253.9 1279.3 1.6521 1.6816 1.7089

110 V 4.1513 (334.78) h 11 98.3 s 1.6068

4.4689 1226.2 1.6402

5.9039 1304.5 1.7344

4.7724 5.0674 5.3568 1252.6 1278.3 1303.6 1.6701 1.6976 1.7233

6.2165 1329.5 1.7586

6.8342 1379.4 1.8037

800

900

7.4456 1429.8 1.8453

1000

1100

1200

1400

10.831 11.583 1532.9 1585.3 1.9457 1.9804

12.333 13.831 1638.5 1747.5 2.0135 2.0755

9.6237 1532.6 1.9326

10.293 1585.0 1.9673

10.960 12.292 1638.3 1747.4 2.0004 2.0625

8.0529 8.6576 1480.7 1532.3 1.8842 1.9209

9.2602 1584.8 19556

9.8615 11.061 1638.1 1747.2 1.9887 2.0508

9.321 6 10.078 1430.6 1481.4 1.8704 1.9092

5.6424 6.2061 6.7634 1328.8 1378.9 1429.4 1.7476 1.7928 1.8345

7.3166 1480.4 1.8735

7.8671 8.41 56 8.9627 10.054 1532.1 1584.6 1637.9 1747.1 1.9102 19450 1.9781 2.0402

5.1640 1328.1 1.7375

5.6827 1378.4 1.7829

6.1 949 1428.9 1.8247

6.7030 1480.0 1.8637

7.2083 1531.8 1.9005

130 V 3.4711 3.7500 (347.33) h 11 93.8 1223.0 s 1.5839 1.6189

4.0130 4.2570 4.5151 4.7592 1250.2 1276.3 1302.0 1327.3 1.6496 1.6776 1.7037 1.7282

5.2398 1377.8 1.7737

5.7138 1428.5 1.8156

6.1 838 1479.7 18547

6.6509 7.1 162 7.5800 8.5047 15315 1584.1 1637.5 1746.8 1.8915 1 9263 19595 2.0217

140 V (353.04) h

3.4673 1221.4 1.6092

3.7145 3.9524 4.1843 4.4122 1248.9 1275.3 1301.1 1326.6 1.6404 1.6687 1.6949 1.7195

4.8602 5.3015 5.7387 6.1732 1377.3 1428.1 1479.3 1531.2 1.7652 1.8072 1.8464 1.8832

150 V (358.43) h

3.2220 3.4557 3.6797 3.8976 1219.7 1247.6 1274.3 1300.3 1.6001 1.6317 1.6602 1.6866

4.1115 4.531 1 4.9441 1325.9 1376.8 1427.7 1.711 4 1.7573 1.7994

160 V (363.55) h

3.0073 3.2291 3.4411 1218.0 1246.3 1273.3 1.5914 1.6235 16523

3.8483 1325.2 1.7038

4.2432 1376.2 1.7498

4.6314 5.0155 5.3968 1427.2 1478.7 1530.7 1.7920 1.8313 1.8682

5.7761 6.1540 6.9069 1583.4 1636.9 1746.4 1.9031 19363 1.9986

170 V (368.43) h

2.8176 3.0291 3.2304 3.4253 3.6160 3.9892 1216.3 1245.0 1272.2 1298.6 1324.5 1375.7 1.5831 1.6157 1.6448 1.6716 1.6966 1.7428

4.3555 4.7177 5.0771 1426.8 1478.3 1530.4 1.7851 1.8244 1.8613

5.4345 5.7905 6.4996 1583.1 1636.7 1746.2 1.8963 1.9296 1.9919

180 V (373.08) h

2.6487 1214.5 15752

2.8512 1243.7 1.6082

V 190 (377.54) h

2.4975 1212.7 1.5676

2.6920 2.8755 1242.4 1270.1 1.601 1 1.6309

200 V (381.81) h

2.3612 2.5485 2.7246 1210.9 1241.0 1269.1 1.5602 1.5943 16243

220 V (389.89) h

2.1252 2.3006 2.4638 2.6198 1207.0 1238.3 1266.9 1294.3 1.5461 15814 1.6121 1.6399

2.7712 3.0652 3.3521 3.6348 1320.8 1373.0 1424.7 1476.6 1.6656 1.7126 1.7554 1 7950

240 V (397.41) h

1.9277 1203.0 1.5327

2.5317 2.8034 1319.4 1371.9 1.6549 1.7023

120 V 3.7832 4.0796 4.3611 (341.26) h 1196.1 1224.6 1251.4 s 1.5950 1.6292 1.6595

V=

specific volume, ft3/1bm

2.0936 1235.4 1.5694

4.6339 1277.3 1.6872

4.9009 1302.8 17131

3.6468 1299.5 1.6789

3.0431 3.2285 1271.2 1297.7 1.6377 16646

2.2462 1264.7 1.6007

h = enthalpy, Btu/lbm

3.4095 1323.8 16898

3.0524 3.2248 1296.9 1323.0 1.6580 1.6833

2.3914 1292.5 1.6289

0

7.8960 1746.7 2.0135

4.7929 5.1309 5.4674 6.1376 1530.1 1582.9 1636.5 1746.1 1.8549 1.8899 19232 19855

3.5613 3.8908 4.2161 4.5387 1374.6 1426.0 1477.6 1529.8 1.7298 1.7723 18118 1.8488

s =entropy, Btu/(1bm· R)

6.6057 7.0368 1583.8 1637.3 19180 19512

5.3530 5.7591 6.1 632 6.5660 7.3685 1479.0 1530.9 1583.6 1637.1 1746.5 1.8386 1.8754 1.9103 19435 2.0058

3.7633 4.11 03 4.4530 1375.1 1426.4 1478.0 1.7361 1.7785 1.8179

2.8938 3.0585 3.3794 3.6933 1296.0 1322.3 1374.1 1425.5 1.6517 1.6771 1.7238 1.7664

7.711 7 8.21 37 9.2148 1584.3 1637.7 1746.9 1 9353 1.9684 2.0306

4.0030 1477.3 1.8059

4.3098 1529.5 1.8430

4.8592 1582.7 1.8838

5.1 784 5.8137 1636.3 1745.9 1.9171 19795

4.6147 1582.4 1.8780

4.9182 5.5222 1636.1 1745.8 1.9114 1.9738

3.9146 4.1924 4.4688 5.0186 1529.0 15819 1635.7 1745.5 1.8322 1.8673 19007 19631

3.0678 3.3279 3.5852 1423.8 1475.9 1528.4 1.7453 1.7851 1.8223

3.8404 1581.5 18575

4.0943 4.5990 1635.3 1745.2 1.8909 1.9534

324  Appendix B Table 3 (continued).

Su~erheated

Pressure psi a (Sat. T)

Temperature-Degrees Fahrenheit

Steam 1300

1400

1500

3.5426 1581.0 1.8484

3.7774 4.0111 1634.9 1689.5 1.8819 1.9138

4.2439 1744.9 1.9445

4.4763 1801.1 1.9739

3.0676 1527.2 1.8047

3.2874 1580.5 1.8400

3.5058 1634.5 1.8735

3.7231 1689.1 1.9055

3.9396 1744.6 1.9362

4.1556 1800.8 1.9656

2.6529 1473.8 1.7593

2.8605 1526.7 1.7968

3.0662 3.2704 1580.0 1634.0 1.8322 1.8657

3.4735 1688.8 1.8978

3.6758 1744.3 1.9284

3.8776 1800.6 1.9579

2.2859 1420.4 1.7115

2.4841 1473.1 1.7518

2.6793 1526.1 1.7894

2.8726 1579.6 1.8248

3.2551 1688.4 1.8905

3.4450 1744.0 1.9212

3.6344 1800.3 1.9507

1.9560 1366.3 1.6603

2.1478 1419.5 1.7043

2.3351 1472.4 1.7447

2.5195 1525.5 1.7824

2.7018 2.8826 1579.1 1633.2 1.8179 1.8516

3.0624 3.2414 1688.1 1743.7 1.8837 1.9144

3.4198 1800.1 1.9439

1.6527 1310.1 1.6036

1.8429 1365.2 1.6533

2.0252 1418.6 1.6975

2.2028 1471.7 1.7381

2.3774 1525.0 1.7758

2.5500 1578.6 1.8114

2.7211 2.8911 1632.8 1687.7 1.8451 1.8772

3.0604 1743.4 1.9080

3.2291 1799.8 1.9375

1.4637 1279.3 1.5683

1.5600 1308.5 1.5965

1.7418 1364.0 1.6466

1.9154 1417.8 1.6910

2.0843 1471.0 1.7317

2.2502 1524.4 1.7696

2.4141 1578.1 1.8052

2.5765 1632.4 1.8389

2.7379 1687.4 1.8711

2.8984 1743.1 1.9019

3.0584 1799.5 1.9314

1.2850 1245.6 1.5288

1.3839 1277.3 1.5611

1.4765 1.6507 1306.9 1362.9 1.5897 1.6402

1.8166 1416.9 1.6848

1.9777 1470.3 1.7257

2.1358 1523.8 1.7636

2.2919 1577.7 1.7993

2.4464 1632.0 1.8331

2.6000 1687.0 1.8653

2.7527 1742.8 1.8961

2.9048 1799.3 1.9257

450 V (456.32) h

1.1232 1238.9 1.5103

1.2151 1272.2 1.5442

1.3001 1302.8 1.5737

1.4584 1360.0 1.6253

1.6079 1414.7 1.6705

1.7526 1.8942 1468.6 1522.4 1.7117 1.7499

2.0337 1576.5 1.7857

2.1718 1631.0 1.8196

2.3088 1686.2 1.8519

2.4449 1742.0 1.8828

2.5805 1798.6 1.9124

500 V (467.05) h

0.9930 1231.9 1.4928

1.0797 1267.0 1.5284

1.1587 1298.6 1.5591

1.3044 1357.0 1.6117

1.4409 1412.4 1.6576

1.5725 1466.8 1.6991

1.7009 1520.9 1.7375

1.8272 1575.3 1.7735

1.9521 1630.0 1.8076

2.0758 1685.3 1.8399

2.1987 1741.3 1.8708

2.3211 1798.0 1.9005

550 V (476.98) h

0.8856 1224.5 1.4760

0.9685 1261.5 1.5137

1.0428 1294.3 1.5454

1.1783 1354.0 1.5993

1.3043 1410.2 1.6457

1.4251 1465.0 1.6876

1.5428 1519.5 1.7263

1.6583 1574.0 1.7624

1.7723 1629.0 1.7966

1.8852 1684.4 1.8290

1.9973 1740.5 1.8600

2.1088 1797.3 1.8898

V 600 (486.25) h

0.7953 1216.5 1.4597

0.8754 1255.8 1.4996

0.9460 1289.9 1.5325

1.0732 1351 .0 1.5877

1.1904 1407.9 1.6348

1.3023 1.4110 1463.2 1518.0 1.6770 1.7159

1.5175 1572.8 1.7523

1.6225 1628.0 1.7865

1.7264 1683.6 1.8190

1.8295 1.9320 1739.8 1796.7 1.8501 1.8799

0.7178 1208.0 1.4434

0.7962 1249.9 1.4861

0.8639 1285.3 1.5203

0.9841 1347.9 1.5768

1.0940 1405.6 1.6246

1.1983 1461.4 1.6672

1.2994 1516.6 1.7063

1.3983 1571 .6 1.7428

1.4957 1626.9 1.7772

1.5920 1682.7 1.8098

1.6874 1739.0 1.8410

1.7823 1796.0 1.8708

07280 1243.7 1.4730

0.7933 1280.6 1.5087

0.9077 1344.8 1.5666

1.0113 1403.3 1.6150

1.1092 1459.6 1.6580

1.2038 1515.1 1.6974

1.2962 1570.4 1.7341

1.3871 1625.9 1.7686

1.4768 1681.8 1.8013

1.5657 1738.3 1.8325

1.6540 1795.4 1.8624

0.6684 1237.3 14602

0.7319 1275.8 1.4975

0.8414 1341.6 1.5569

0.9396 1401.0 1.6060

1.0320 1457.8 1.6494

1.1209 1513.6 1.6890

1.2077 1569.2 1.7259

1.2929 1624.9 1.7605

1.3770 1681.0 1.7933

1.4602 1.5428 1737.5 1794.7 1.8245 1.8545

450

500

550

600

700

800

900

1000

11 00

1.9182 1232.5 1.5580

2.0620 1262.5 1.5901

2.1980 1290.7 1.6188

2.3290 1317.9 1.6450

2.5818 1370.8 1.6928

2.8272 1423.0 1.7359

3.0683 1475.2 1.7758

3.3065 1527.8 1.8132

1.7676 1229.5 1.5473

1.9039 1260.2 1.5801

2.0322 1288.8 1.6092

2.1552 1316.3 1.6358

2.3919 1369.7 1.6839

2.6210 1422.1 1.7273

2.8457 1474.5 1.7673

1.6367 1226.4 1.5370

1.7668 1257.9 1.5706

1.8883 1287.0 1.6002

2.0045 1314.8 1.6271

2.2272 1368.5 1.6756

2.4423 1421.2 1.7191

1.5219 1223.3 1.5271

1.6467 1255.5 1.5615

1.7624 1285.1 1.5916

1.8726 1313.3 1.6189

2.0831 1367.4 1.6677

1.4203 1220.0 1.5175

1.5405 1253.1 1.5529

1.6512 1283.2 1.5835

1.7562 1311.7 1.6111

V 1.3297 360 (434.43) h 1216.6 s 1.5082

1.4460 1250.6 1.5446

1.5523 1281 .3 1.5757

V 1.2483 380 (439.63) h 1213.1 s 1.4991

1.3613 1248.1 1.5365

260

V

(404.45) h

s

280

V

(411.09) h

s

300

V

(417.37) h

s

320

V

(423.33) h

s

340

V

(429.01) h

s

400

V

(444.63) h

s

650

V

(494.94) h

1.1747 1209.5 1.4901

700 V (503.14) h 750

V

(510.90) h

V=specific volume, ft3Jibm

h=enthalpy, Btu/lbm

s =entropy, Btu/(lbm·0 R)

1200

3.0644 1633.6 1.8584

Steam Tables  325 Table 3 (continued}.

Su~erheated

Pressure psi a (Sat. T)

Temperature-Degrees Fahrenheit

Steam

550

600

650

700

750

800

900

800 V 0.6159 (518.27) h 1230.5 s 1.4476

0.6780 1270.8 1.4866

0.7328 1306.0 1.5191

0.7834 1338.4 1.5476

0.8311 1369.0 1.5735

0.8768 1398.6 1.5975

850 V 0.5691 (525.30) h 1223.4 s 1.4351

0.6302 1265.7 1.4761

0.6834 1302.0 1.5096

0.7320 1335.1 1.5388

07777 1366.3 1.5651

0.5269 0.5875 1215.8 ' 1260.4 1.4226 1.4658

0.6394 1297.9 1.5004

0.6864 1331 .8 1.5302

0.4887 1207.8 1.4100

0.5491 1254.9 1.4557

0.5998 1293.7 1.4914

0.4538 1199.1 1.3971

0.5143 1249.3 1.4457

900

V

(532.02) h

s

950

V

(538.46) h

s

1000

V

(544.65) h

s

1100

V

1200

V

1300

V

1400

V

1500

V

1600

V

1700

V

1800

V

1900

V

2000

V

(556.35) h

(567.26) h

(577.50) h

(587.14) h

(596.27) h

1100

1200

1300

1400

1500

0.9643 1456.0 1.6413

1.0484 1.1302 1512.1 1568.0 1.6812 1.7182

1.2105 1623.9 1.7529

1.2896 1680.1 1.7858

1.3679 1736.8 1.8171

1.4456 1794.0 1.8471

0.8214 1396.2 1.5894

0.9047 1454.1 1.6336

0.9844 1510.7 1.6737

1.0619 1566.7 1.7109

1.1378 1622.8 1.7457

1.2125 1679.2 17787

1.2864 1736.0 1.8101

1.3597 1793.4 1.8401

0.7303 1363.5 1.5570

0.7721 1393.8 1.5816

0.8516 1452.3 1.6262

0.9275 1509.2 1.6666

1.0011 1565.5 1.7040

1.0731 1621 .8 1.7389

1.1440 1678.3 1.7720

1.2140 1735.3 1.8035

1.2834 1792.7 1.8336

0.6454 1328.4 1.5220

0.6878 1360.6 1.5492

0.7280 1391.4 1.5742

0.8041 1450.4 1.6192

0.8766 1507.7 1.6599

0.9467 1564.3 1.6974

1.0153 1620.8 1.7325

1.0827 1677.4 1.7657

1.1492 1734.5 1.7972

1.2152 1792.1 1.8273

0.5641 1289.4 1.4827

0.6085 1324.9 1.5140

0.6495 1357.8 1.5418

0.6883 1389.0 1.5670

0.7614 1448.5 1.6125

0.8308 1506.2 1.6535

0.8978 1563.0 1.6911

0.9632 1619.7 1.7264

1.0275 1676.6 1.7596

1.0909 1733.7 1.7912

1.1537 1791.4 1.8214

0.4536 1237.2 1.4259

0.5022 1280.5 1.4658

0.5446 1317.9 1.4987

0.5833 1351.9 1.5275

0.6196 1384.0 1.5535

0.6875 1444.7 1.5999

0.7516 1503.2 1.6414

0.8133 1560.6 1.6794

0.8733 1617.7 1.7149

0.9322 1674.8 1.7483

0.9902 1732.2 1.7801

1.0476 1790.1 1.8104

0.4020 1224.1 1.4061

0.4501 1271.1 1.4494

0.4910 1310.5 1.4842

0.5279 1345.9 1.5141

0.5622 1378.9 1.5408

0.6259 1440.9 1.5882

0.6856 1500.1 1.6302

0.7428 1558.1 1.6686

0.7984 1615.6 1.7043

0.8528 1673.0 1.7379

0.9063 1730.7 1.7698

0.9592 1788.8 1.8002

0.3574 1209.8 1.3859

0.4057 1261.3 1.4334

0.4455 1302.9 1.4701

0.4809 1339.7 1.5012

0.5136 1373.7 1.5288

0.5738 1437.0 1.5772

0.6298 1497.0 1.6198

0.6832 1555.6 1.6585

0.7350 1613.5 1.6945

0.7856 1671.3 1.7283

0.8353 1729.2 1.7604

0.8843 1787.4 1.7909

0.3178 1193.7 1.3649

0.3671 1250.8 1.4175

0.4063 1295.0 1.4566

0.4405 1333.4 1.4890

0.4718 1368.5 1.5174

0.5290 1433.1 1.5668

0.5819 1493.9 1.6100

0.6321 1553.1 1.6491

0.6806 1611.4 1.6854

0.7280 1669.5 1.7194

07744 1727.7 1.7515

0.8202 1786.1 1.7821

0.2819 1175.4 1.3423

0.3331 1239.6 1.4016

0.3720 1286.8 1.4433

0.4054 1326.9 1.4771

0.4356 1363.1 1.5064

0.4902 1429.1 1.5569

0.5403 1490.8 1.6007

0.5878 1550.5 1.6403

0.6335 1609.3 1.6768

0.6780 1667.7 1.7110

0.7217 1726.1 1.7433

0.7646 1784.8 1.7740

0.3029 1227.7 1.3855

0.3418 1278.3 1.4302

0.3745 1320.2 1.4656

0.4037 1357.6 1.4959

0.4562 1425.1 1.5475

0.5040 1487.7 1.5920

0.5490 1548.0 1.6319

0.5923 1607.2 1.6687

0.6343 1665.9 1.7031

0.6755 1724.6 1.7355

0.7160 1783.5 1.7663

0.2757 1214.7 1.3691

0.3149 1269.3 1.4172

0.3471 1313.3 1.4544

0.3756 1351.9 1.4857

0.4262 1421.0 1.5385

0.4719 1484.5 1.5836

0.5148 1545.4 1.6240

0.5559 1605.0 1.6610

0.5958 1664.1 1.6956

0.6348 1723.0 1.7282

0.6731 1782.1 1.7591

0.2507 1200.6 1.3520

0.2908 1259 9 1.4043

0.3227 1306.2 1.4434

0.3505 1346.2 1.4758

0.3994 14169 1.5299

0.4433 1481 .3 1.5756

0.4844 1542.8 1.6164

0.5236 16029 1.6537

0.5615 1662.3 1.6885

0.5986 1721 .5 1.7212

0.6349 1780.8 1.7522

0.2277 1185.1 1.3340

0.2689 1250.1 1.3914

0.3007 1298.8 1.4325

0.3280 1340.3 1.4662

0.3755 1412.7 1.5215

0.4178 1478.1 1.5680

0.4572 1540.2 1.6091

0.4947 1600.8 1.6468

0.5309 1660.5 1.6817

0.5662 1720.0 1.7146

0.6008 1779.5 1.7457

0.2059 1167.5 1.3146

0.2489 1239.7 1.3783

0.2807 0.3076 1291 .2 1334.3 1.4218 1.4567

0.3539 1408.5 1.5134

0.3948 1474.9 1.5606

0.4327 0.4686 1537.6 1598.6 1.6022 1.6401

0.5033 1658.7 1.6752

0.5370 1718.4 1.7083

0.5701 1778.1 1.7395

(604.93) h

(613.19) h

(621.07) h

(628.62) h

(635.85) h

V=

specific volume, ft3/1bm

h = enthalpy, Btu/lbm

s=entropy, Btu/(1bm· R) 0

1000

326  Appendix B Table 3 (continued}. Table 3 (continued). Pressure Pressure psi a psi a (Sat. T) (Sat T)

Su~erheated Su~erheated

Temperature-Degrees Fahrenheit Temperature550 600 Degrees 700 650Fahrenheit

Steam Steam

650

700

750

800

750 850

800 V 0.6159 2200 hV 1230.5 0.1635 (518.27) (649.50) sh 1.4476 1122.0 s 1.2673

0.6780 0.2136 1270.8 1217.1 1.4866 1.3514

0.7328 0.2459 1306.0 1275.2 1.5191 1.4006

0.7834 0.2723 1338.4 1321 .8 1.5476 1.4383

850 V 0.5691 2400 hV 1223.4 (525.30) (662.16) h 1.4351 s

0.6302 0.1827 1265.7 1191.0 1.4761 1.3226

0.6834 0.2164 1302.0 1258.0 1.5096 1.3793

900 V 0.5269 0.5875 2600 V 0.1548 (532.02) h 1215.8 ' 1260.4 (67398) h 1160.0 s 1.4226 1.4658 1.2905

800 900

900 950

1000 1000

1100 1100

1200 1200

1300 1300

1400 1400

1500 1500

0.8311 0.2955 1369.0 1362.6 1.5735 1.4700

0.8768 0.9643 1.0484 0.3166 1512.1 1456.0 0.3550 1398.6 0.3363 1399.8 1.5975 1434.8 1.6413 1468.3 1.6812 1.4980 1.5232 1.5466

1.1302 0.3903 1568.0 1532.4 1.7182 1.5891

0.7320 0.2426 1335.1 1308.8 1.5388 1.4204

07777 0.2651 1366.3 1352.0 1.5651 1.4541

0.8214 0.9047 0.9844 1.0619 1.1378 1.2125 1.2864 1.3597 0.2854 0.3042 0.3219 1566.7 0.3861 1679.2 0.4447 1736.0 0.4729 1793.4 1396.2 1454.1 1510.7 0.3550 1622.8 0.4159 1427.2 1461 .6 1527.1 1.7457 1651.4 1712.2 1772.8 1390.9 1.5894 1.6336 1.6737 1.7109 1589.9 17787 1.8101 1.8401 1.4833 1.5094 1.5335 1.5769 1.6159 1.6519 1.6855 1.7172

0.6394 0.1909 1297.9 1239.4 1.5004 1.3577

0.6864 0.2173 1331 .8 1295.0 1.5302 1.4027

0.7303 0.2393 1363.5 1341.0 1.5570 1.4386

0.7721 0.8516 0.9275 0.2589 0.2769 0.2938 1393.8 1452.3 1509.2 1381.8 1419.4 1454.8 1.5816 1.6262 1.6666 1.4692 1.4963 1.5210

950 V 0.4887 2800 V (538.46) h 1207.8 (685.03) h s 1.4100

0.5491 0.5998 0.1280 0.1685 1254.9 1293.7 1120.6 1219.0 1.4557 1.4914 1.2520 1.3353

0.6454 0.1953 1328.4 1280.5 1.5220 1.3852

0.6878 0.2171 1360.6 1329.7 1.5492 1.4235

0.7280 0.8041 0.8766 0.9467 1.0153 0.2362 0.2535 0.2697 0.2995 0.3272 1391.4 1450.4 1507.7 1564.3 1620.8 1372.5 1411.5 1447.9 1516.3 1581.2 1.5742 1.6192 1.6599 1.6974 1.7325 1.4555 1.4837 1.5092 1.5545 1.5948

1000 3000 VV 0.4538 (544.65) h 1199.1 (695.41) h s 1.3971

0.5143 0.0984 1249.3 1059.8 1.4457 1.1959

0.5641 0.1484 1289.4 1196.4 1.4827 13118

0.6085 0.1760 1324.9 1265.2 1.5140 1.3675

0.6495 0.1977 1357.8 1317.9 1.5418 1.4086

0.6883 0.2332 0.7614 0.2164 1389.0 1403.4 1448.5 1362.9 1.5670 1.4716 1.6125 1.4423

1.2105 1.2896 1.3679 1.4456 0.4236 1623.9 0.4556 1680.1 0.4867 1736.8 0.5171 1794.0 1594.3 1.7529 1655.1 1.7858 1715.3 1.8171 1775.5 1.8471 1.6276 1.6631 1.6964 1.7279

1.0011 1.0731 1.1440 1.2140 0.3251 0.3544 0.3823 0.4092 1565.5 1621 .8 1678.3 1735.3 1521.7 1585.6 1647.8 1709.1 1.7040 1.7389 1.7720 1.8035 1.5654 1.6050 1.6414 1.6753

1.2834 0.4355 1792.7 1770.1 1.8336 1.7073

1.0827 1.1492 1.2152 0.3535 0.3788 0.4034 1677.4 1734.5 1792.1 1644.1 1706.0 1767.4 1.7657 1.7972 1.8273 1.6316 1.6658 1.6980

0.8308 0.8978 0.9632 1.0275 1.0909 0.3757 0.2487 0.2773 0.3037 0.3286 0.3525 1.1537 1506.2 1510.9 1563.0 1576.7 1619.7 1640.4 1676.6 1702.8 1733.7 1764.7 1791.4 1441.0 1.8214 1.6535 1.5442 1.6911 1.5851 1.7264 1.6223 1.7596 1.6569 1.7912 1.6893 1.4978

1100 3200

V V

0.4536 1237.2 1.4259

0.5022 0.1300 1280.5 1171.0 1.4658 1.2866

0.5446 0.1589 1317.9 1248.9 1.4987 1.3497

0.5833 0.1806 1351.9 1305.7 1.5275 1.3939

0.8733 0.3067 0.9322 0.3294 0.9902 0.3514 1.0476 0.6196 0.2154 0.6875 0.2304 0.7516 0.2579 0.8133 0.2830 0.1990 1617.7 1636.7 1674.8 1699.7 1732.2 1762.0 1790.1 1384.0 1395.1 1444.7 14339 1503.2 1505.4 1560.6 1572.3 1353.0 1.5535 1.4598 1.5999 1.4869 1.6414 1.5343 1.6794 1.5759 1.7149 1.6136 1.7483 1.6484 1.7801 1.6810 1.8104 14294

1200 3400

V V

0.4020 1224.1 1.4061

0.4501 0.1129 1271.1 1141.8 1.4494 1.2587

0.4910 0.1435 1310.5 1231.6 1.4842 13316

0.5279 0.1654 1345.9 1292.9 1.5141 1.3793

0.5622 0.1996 0.6259 0.2143 0.6856 0.2407 0.7428 0.2648 0.7984 0.2875 0.8528 0.1836 1673.0 1378.9 1386.7 1440.9 1426.6 1500.1 1499.8 1558.1 1567.8 1615.6 16330 1342.9 1.5408 1.4484 1.5882 1.4763 1.6302 1.5248 1.6686 1.5671 1.7043 1.6052 1.7379 1.4168

0.9063 0.3091 1730.7 1696.6 1.7698 1.6403

0.9592 0.3299 1788.8 1759.3 1.8002 1.6732

1300 3600

VV

0.3574 1209.8 1.3859

0.4057 0.0964 1261.3 1107.2 1.4334 1.2269

0.4455 0.1296 1302.9 1213.2 1.4701 1.3129

0.4809 0.1518 1339.7 1279.7 1.5012 1.3648

0.7856 0.2910 0.5136 0.1856 0.5738 0.1999 0.6298 0.2255 0.6832 0.2487 0.8353 0.7350 0.2704 0.1699 1729.2 1373.7 1378.1 1437.0 1419.3 1497.0 1494.2 1555.6 1563.3 1613.5 1629.2 1671.3 1693.4 1332.5 1.5288 1.4373 1.5772 1.4660 1.6198 1.5157 1.6585 1.5586 1.6945 1.5972 1.7283 1.6327 1.7604 1.4043

0.8843 0.3109 1787.4 1756.6 1.7909 1.6658

1400 3800

VV

0.3178 1193.7 1.3649

0.3671 0.0802 1250.8 1064.4 1.4175 1.1888

0.4063 0.1169 1295.0 1193.4 1.4566 1.2936

0.4405 0.1396 1333.4 1266.0 1.4890 1.3502

0.4718 0.1576 1368.5 1321.9 1.5174 1.3921

0.5290 0.1870 0.5819 0.2118 0.6321 0.1731 1433.1 1411.9 1493.9 1488.5 1553.1 1369.4 1.5668 1.4561 1.6100 1.5069 1.6491 1.4264

1500 4000

VV

0.2819 1175.4 1.3423

0.3331 0.0637 1239.6 1009.2 1.4016 1.1409

0.3720 0.1052 1286.8 1172.1 1.4433 1.2734

0.4054 0.1285 1326.9 1251.7 1.4771 1.3355

0.4356 0.1465 1363.1 1310.9 1.5064 1.3799

0.4902 0.1754 0.5403 0.1996 0.5878 0.2212 0.6335 0.2413 0.6780 0.2603 0.7217 0.1617 1429.1 1404.4 1490.8 1482.8 1550.5 1554.2 1609.3 1621.7 1667.7 1687.1 1726.1 1360.5 1.5569 1.6007 1.6403 1.6768 1.7110 1.7433 14157 14463 1.4983 1.5427 1.5822 1.6183

0.7646 0.2785 1784.8 1751.2 1.7740 1.6519

1600 4500

VV

0.3029 0.0393 1227.7 891.0 1.3855 1.0395

0.3418 0.0796 1278.3 1111.1 1.4302 1.2183

0.3745 0.1047 1320.2 1213.4 1.4656 1.2980

0.4037 0.1229 1357.6 1282.3 1.4959 1.3497

0.4562 0.1509 0.5040 0.1737 0.5490 0.1938 0.5923 0.2122 0.6343 0.2296 0.6755 0.1378 1425.1 1385.3 1487.7 1468.4 1548.0 1542.7 1607.2 1612.3 1665.9 1679.2 1724.6 1337.5 1.5475 1.4229 1.5920 1.4780 1.6319 1.5242 1.7031 1.6019 1.6687 1.5650 1.7355 1.3896

0.7160 0.2462 1783.5 1744.5 1.7663 1.6361

1700 5000

VV

0.2757 0.0337 1214.7 8530 1.3691 1.0053

0.3149 0.0594 1269.3 1041.9 1.4172 1.1582

0.3471 0.0855 1313.3 1171.5 1.4544 1.2593

0.3756 0.1039 1351.9 1252.1 1.4857 13198

0.4262 0.1313 0.4719 0.1530 0.6731 0.6348 0.2204 0.5148 0.1719 0.5559 0.1890 0.5958 0.2051 0.1186 1782.1 1421.0 1365.5 1484.5 14538 1545.4 1531.2 1605.0 1602.9 1664.1 16713 1723.0 1737.7 1313.7 1.5385 1.4005 1.5836 1.4590 1.6240 1.5071 1.6956 1.5869 1.7282 1.6217 1.7591 1.6610 1.5491 1.3643

1800 5500

VV

0.2507 0.0313 0.2908 0.0463 1200.6 834.1 1259 980.99 1.3520 0.9872 1.4043 1.1060

0.3227 0.0701 1306.2 1126.9 1.4434 1.2198

0.3505 0.0885 1346.2 1220.4 1.4758 1.2899

0.3994 0.1030 14169 1289.1 1.5299 1.3396

0.4433 0.1361 0.4844 0.1540 0.5236 0.1701 0.5615 0.1851 0.5986 0.1993 0.6349 0.1153 1481 .3 1439.0 1721 .5 1731.0 1780.8 1542.8 1519.6 16029 1593.4 1662.3 1663.4 1345.4 1.7212 1.6084 1.7522 1.5756 14409 1.6164 1.4910 1.6537 1.5343 1.6885 1.5729 1.3788

1900 6000

VV

0.2277 0.0298 1185.1 821.7 0.9747 1.3340

0.2689 0.0395 1250.1 940.8 1.0710 1.3914

0.3007 0.0582 10831 1298.8 1.1818 1.4325

0.3280 0.0759 1187.7 1340.3 1.2604 1.4662

0.3755 0.0901 12638 1412.7 1.3154 1.5215

0.4572 0.1391 0.4947 0.1544 0.5309 0.1684 0.5662 0.1817 0.6008 0.1021 0.4178 0.1221 1779.5 1324.8 1478.1 1424.0 1540.2 1507.9 1600.8 1583.9 1660.5 1655.5 1720.0 1724.3 1.6091 1.4759 1.6468 1.5204 1.6817 1.5599 1.7146 1.5960 1.7457 1.3579 1.5680 1.4237

2000 7000

VV

0.2059 0.0279 805.6 1167.5 0.9570 1.3146

0.0334 0.2489 898.4 1239.7 1.0321 1.3783

0.0438 0.3076 0.0576 0.2807 1124.8 1013.3 1291 .2 1334.3 1.1215 1.4567 1.2051 1.4218

0.0705 0.3539 1213.1 1408.5 1.2689 1.5134

0.0817 0.3948 0.1004 0.4327 0.1160 0.4686 0.1298 0.5033 0.1424 0.5370 0.1542 0.5701 1283.4 1778.1 1474.9 1394.0 1537.6 1484.6 1598.6 1565.1 1658.7 1639.8 1718.4 1711.0 1.3179 1.5606 1.3913 1.6022 1.4476 1.6401 1.4948 1.6752 1.5361 1.7083 1.5734 1.7395

(556.35) (705.10) hh

(567.26) hh

(577.50) hh

(587.14) hh

(596.27) hh

(604.93) hh

(613.19) hh

(621.07) hh

(628.62) hh

(635.85) hh

v=specific specificvolume, volume,ft3/1bm ft3/1bm V=

enthalpy,Btu/lbm Btu/lbm hh==enthalpy,

entropy, Btu/(1bm· Btu/(1bm· R) R) sS==entropy, 00

07744 0.2939 0.6806 0.2551 0.7280 0.2748 0.8202 0.2342 1611.4 1625.5 1669.5 1690.3 1727.7 17539 1786.1 1558.8 1.6854 1.5896 1.7194 1.6254 1.7515 1.6587 1.7821 1.5505

Steam Tables  327 Table 3 {continued).

Su~erheated

Pressure psi a (Sat. T)

8000

Temperature-Degrees Fahrenheit V

h

s

9000

V

h

s

10000

V

h

s

11000

V

h

s

12000

V

h

s

13000

V

h

s

14000

V

h

s

15000

V

h

s

V=

Steam

750

800

850

900

950

1000

1050

1100

1150

0.0267 795.1 0.9441

0.0306 876.0 1.0096

0.0371 971.0 1.0836

0.0465 1073.2 1.1601

0.0571 1165.6 1.2269

0.0672 1243.0 1.2808

0.0763 1307.9 1.3246

0.0844 1364.3 1.3614

0.0258 787.5 0.9338

0.0289 861.6 0.9938

0.0335 945.1 1.0588

0.0401 1036.2 1.1270

0.0483 0.0568 0.0650 1125.7 1205.8 1275.1 1.1917 ·1.2475 1.2942

0.0251 0.0276 0.0312 781.8 • 851 .3 927.6 0.9252 0.9815 1.0409

0.0362 1010.1 1.1027

0.0425 1094.2 1.1634

0.0495 1173.6 1.2188

0.0245 777.3 0.9177

0.0267 843.6 0.9714

0.0296 915.0 1.0269

0.0336 991.3 1.0841

0.0385 1070.0 1.1409

0.0240 773.8 0.9111

0.0260 837.5 0.9627

0.0285 905.3 1.0155

0.0317 977.1 1.0692

0.0236 771.0 0.9051

0.0253 832.7 0.9551

0.0275 897.8 1.0057

0.0232 768.7 0.8996

0.0248 828.8 0.9483

0.0229 0.0243 766.9 825.6 0.8946 0.9422

specific volume, ft3/1bm

1300

1400

1500

0.0919 1414.9 1.3933

0.0988 0.1115 1461 .5 1546.4 1.4218 1.4715

0.1230 1624.3 1.5146

0.1337 1697.9 1.5532

0.0725 1335.6 1.3337

0.0794 1389.6 1.3678

0.0858 1438.9 1.3980

0.0975 1528.1 1.4502

0.1081 1609.1 1.4949

0.1179 1685.1 1.5348

0.0566 1245.0 1.2669

0.0633 1308.5 1.3083

0.0697 1365.4 1.3442

0.0756 1417.2 1.3759

0.0864 1510.4 1.4304

0.0962 1594.3 1.4768

0.1053 1672.6 1.5179

0.0442 1146.8 1.1945

0.0503 1218.5 1.2428

0.0563 1283.5 1.2852

0.0621 1342.6 1.3225

0.0675 1396.5 1.3555

0.0776 1493.3 1.4121

0.0867 1579.9 1.4600

0.0952 1660.4 1.5022

0.0357 1051.3 1.1228

0.0404 1125.1 1.1743

0.0456 0.0509 1195.6 1261.2 1.2218 1.2646

0.0561 1321.6 1.3027

0.0611 1377.1 1.3366

0.0704 1477.0 1.3951

0.0789 1566.1 1.4444

0.0868 1648.6 1.4876

0.0303 966.1 1.0569

0.0336 1036.6 1.1079

0.0376 1107.5 1.1573

0.0420 1176.3 1.2036

0.0467 1241.6 1.2462

0.0513 1302.6 1.2847

0.0559 1359.2 1.3193

0.0645 1461.5 1.3792

0.0724 1552.9 1.4297

0.0798 1637.2 1.4739

0.0267 891.7 0.9973

0.0291 957.3 1.0464

0.0320 0.0354 1024.9 1093.1 1.0952 1.1428

0.0392 0.0433 1160.1 1224.6 1.1879 1.2299

0.0475 1285.7 1.2685

0.0516 1342.8 1.3035

0.0596 1446.9 1.3644

0.0670 1540.3 1.4161

0.0739 1626.3 1.4611

0.0261 886.7 0.9897

0.0282 950.1 1.0372

0.0308 1015.3 1.0843

0.0370 1146.5 1.1742

0.0443 1270.6 1.2539

0.0481 1327.9 1.2890

0.0554 1433.3 1.3506

0.0624 1528.4 1.4032

0.0689 1615.8 1.4490

h = enthalpy, Btu/lbm

0.0337 1081.2 1.1303

s =entropy, Btu/(1bm· R) 0

0.0406 1209.8 1.2155

1200

328  Appendix B Table 4. Properties of Saturated Water and Steam (Temperature) Temp.

"C

Pressure MP a

Volume, m3jkg VL

Vv

Enthalpy, kJ/kg hl hv

Entropy, kJ/(kg·K)

Temp.

oc

~

Sv

2500.9 2510.1 2519.2 2528.4 2537.5

0.0000 0.0763 0.1511 0.2245 0.2965

9.1555 9.0249 8.8998 8.7804 8.6661

0.01 5 10 15 20

104.84 125.75 146.64 167.54 188.44

2546.5 2555.6 2564.6 2573.5 2582.5

0.3673 0.4368 0.5052 0.5724 0.6386

8.5568 8.4521 8.3518 8.2557 8.1634

25 30 35 40 45

0.01 5 10 15 20

0.0006117 0.0008726 0.001228 0.001706 0.002339

0.0010002 0.0010001 0.0010003 0.0010009 0.0010018

206.00 147.02 106.31 77.881 57.761

25 30 35 40 45

0.003170 0.004247 0.005629 0.007384 0.009594

0.0010030 0.0010044 0.0010060 0.0010079 0.0010099

43.341 32.882 25.208 19.517 15.253

50 55 60 65 70

0.012351 0.015761 0.019946 0.025041 0.031201

0.0010121 12.028 0.0010145 9.5649 0.0010171 7.6677 0.0010199 6.1938 0.0010228 5.0397

209.34 230.24 251.15 272.08 293.02

2591.3 2600.1 2608.8 2617.5 2626.1

0.7038 0.7680 0.8312 0.8935 0.9550

8.0749 7.9899 7.9082 7.8296 7.7540

50 55 60 65 70

75 80 85 90 95

0.038595 0.047415 0.057867 0.070182 0.084609

0.0010258 0.0010290 0.0010324 0.0010359 0.0010396

4.1291 3.4053 2.8259 2.3591 1.9806

313.97 334.95 355.95 376.97 398.02

2634.6 2643.0 2651.3 2659.5 2667.6

1.0156 1.0754 1.1344 1.1927 1.2502

7.6812 7.6110 7.5434 7.4781 7.4150

75 80 85 90 95

100 105 110 115 120

0.10142 0.12090 0.14338 0.1 6918 0.19867

0.0010435 0.0010474 0.0010516 0.0010559 0.0010603

1.6719 1.4185 1.2094 1.0359 0.89130

419.10 440.21 461.36 482.55 503.78

2675.6 2683.4 2691.1 2698.6 2705.9

1.3070 1.3632 1.4187 1.4735 1.5278

7.3541 7.2951 7.2380 7.1827 7.1291

100 105 110 115 120

125 130 135 140 145

0.23222 0.27026 0.31320 0.36150 0.41563

0.0010649 0.0010697 0.0010747 0.0010798 0.0010850

0.77011 0.66808 0.58180 0.50852 0.44602

525.06 546.39 567.77 589.20 610.69

2713.1 2720.1 2726.9 2733.4 2739.8

1.5815 1.6346 1.6872 1.7393 1.7909

7.0770 7.0264 6.9772 6.9293 6.8826

125 130 135 140 145

150 155 160 165 170

0.47610 0.54342 0.61814 0.70082 0.79205

0.0010905 0.0010962 0.0011020 0.0011080 0.0011143

0.39250 0.34650 0.30682 0.27246 0.24262

632.25 653.88 675.57 697.35 719.21

2745.9 2751.8 2757.4 2762.8 2767.9

1.8420 1.8926 1.9428 1.9926 2.0419

6.8370 6.7926 6.7491 6.7066 6.6649

150 155 160 165 170

175 180 185 190 195

0.89245 1.0026 1.1233 1.2550 1.3986

0.0011207 0.0011274 0.0011343 0.0011414 0.0011488

0.21660 0.19386 0.17392 0.15638 0.14091

741.15 763.19 785.32 807.57 829.92

2772.7 2777.2 2781.4 2785.3 2788.9

2.0909 2.1395 2.1878 2.2358 2.2834

6.6241 6.5841 6.5447 6.5060 6.4679

175 180 185 190 195

200 205 210 215 220

1.5547 1.7240 1.9074 2.1055 2.3193

0.0011565 0.0011645 0.0011727 0.0011813 0.0011902

0.12722 0.11509 0.10430 0.094689 0.086101

852.39 874.99 897.73 920.61 943.64

2792.1 2794.9 2797.4 2799.4 2801 .1

2.3308 2.3779 2.4248 2.4714 2.5178

6.4303 6.3932 6.3565 6.3202 6.2842

200 205 210 215 220

0.001 21.019 42.021 62.984 83.920

Steam Tables  329 Table 4 . Properties of Saturated Water and Steam (Temperature) Temp.

oc

Pressure MPa

Volume, m3jkg VL

Vv

Enthalpy, kJ/ kg hl

Entropy, kJ/(kg ·K)

Temp.

hv

St.

Sv

oc

225 230 235 240 245

2.5494 2.7968 3.0622 3.3467 3.6509

0.001 199 0.001209 0.001219 0.001229 0.001240

0.078411 0.071510 0.065304 0.059710 0.054658

966.84 990.21 1013.8 1037.5 1061.5

2802.3 2803.0 2803.3 2803.1 2802.3

2.5641 2.6102 2.6561 2.7019 2.7477

6.2485 6.2131 6.1777 6.1425 6.1074

225 230 235 240 245

250 255 260 265 270

3.9759 4.3227 4.6921 5.0851 5.5028

0.001252 0.001264 0.001276 0.001289 0.001303

0.050087 0.045941 0.042175 0.038748 0.035622

1085.7 1110.1 1134.8 1159.8 1185.1

2801.0 2799.1 2796.6 2793.5 2789.7

2.7934 2.8391 2.8847 2.9304 2.9762

6.0722 60370 6.0017 5.9662 5.9304

250 255 260 265 270

275 280 285 290 295

5.9463 6.4165 6.9145 7.4416 7.9990

0.001318 0.001333 0.001349 0.001366 0.001385

0.032767 0.030154 0.027758 0.025557 0.023531

1210.7 1236.7 1263.0 1289.8 1317.0

2785.1 2779.8 2773.7 2766.6 2758.6

3.0221 3.0681 3.1143 3.1608 3.2076

5.8943 5.8578 5.8208 5.7832 5.7449

275 280 285 290 295

300 305 310 315 320

8.5877 9.2092 9.8647 10.556 11 .284

0.001404 0.001425 0.001448 0.001472 0.001499

0.021663 0.019937 0.018339 0.016856 0.015476

1344.8 1373.1 1402.0 1431.6 1462.1

2749.6 2739.4 2727.9 2715.1 2700.7

3.2547 3.3024 3.3506 3.3994 3.4491

5.7058 5.6656 5.6243 5.5816 5.5373

300 305 310 315 320

325 330 335 340 345

12.051 12.858 13.707 14.600 15.540

0.001528 0.001561 0.001597 0.001638 0.001685

0.014189 0.012984 0.011852 0.010784 0.009770

1493.4 1525.7 1559.3 1594.4 1631.4

2684.5 2666.2 2645.6 2622.1 2595.0

3.4997 3.5516 3.6048 3.6599 3.7175

5.4911 5.4425 5.3910 5.3359 5.2763

325 330 335 340 345

350 355 360 365 370 Tc

16.529 17.570 18.666 19.822 21.043 22.064

0.001740 0.001808 0.001895 0.002016 0.002222 0.00311

0.008801 0.007866 0.006945 0.006004 0.004946 0.00311

1670.9 1713.7 1761 .5 1817.6 1892.6 2087.5

2563.6 2526.4 2481.0 2422.0 2333.5 2087.5

3.7783 3.8438 3.9164 4.0011 4.1142 4.4120

5.2109 5.1377 5.0527 4.9482 4.7996 4.4120

350 355 360 365 370 Tc

fc = 373.946 oc

330  Appendix B Table 5. Properties of Saturated Water and Steam (Pressure) Press.

Temp.

MPa

oc

0.001 0.002 0.003 0.004 0.005

6.97 17.50 24.08 28.96 32.88

0.006 0.007 0.008 0.009 0.010

Volume, m3j kg

Enthalpy, kJ / kg

Entropy, kJ/(kg·K)

Press.

hl

hv

SL

Sv

0.0010001 129.18 0.0010014 66.990 0.0010028 45.655 0.0010041 34.792 0.0010053 28.186

29.298 73.435 100.99 121.40 137.77

2513.7 2532.9 2544.9 2553.7 2560.8

0.1059 0.2606 0.3543 0.4224 0.4763

8.9749 8.7227 8.5766 8.4735 8.3939

0.001 0.002 0.003 0.004 0.005

36.16 39.00 41.51 43.76 45.81

0.0010064 0.0010075 0.0010085 0.0010094 0.0010103

151.49 163.37 173.85 183.26 191 .81

2566.7 2571.8 2576.2 2580.3 2583.9

0.5209 0.5591 0.5925 0.6223 0.6492

8.3291 8.2746 8.2274 8.1859 8.1489

0.006 0.007 0.008 0.009 0.010

0.012 0.014 0.016 0.018 0.020

49.42 52.55 55.31 57.80 60.06

0.00101 19 12.359 0.0010133 10.691 0.0010147 9.4309 0.0010160 8.4433 0.0010171 7.6482

206.91 219.99 231 .55 241 .95 251.40

2590.3 2595.8 2600.7 2605.0 2608.9

0.6963 0.7366 0.7720 0.8035 0.8320

8.0850 8.0312 7.9847 7.9437 7.9072

0.012 0.014 0.016 0.018 0.020

0.025 0.030 0.035 0.040 0.045

64.96 69.10 72.68 75.86 78.71

0.0010198 0.0010222 0.0010244 0.0010264 0.0010282

6.2034 5.2286 4.5252 3.9931 3.5761

271.93 289.23 304.25 317.57 329.55

2617.4 2624.6 2630.7 2636.1 2640.9

0.8931 0.9439 0.9876 1.0259 1.0601

7.8302 7.7675 7.7146 7.6690 7.6288

0.025 0.030 0.035 0.040 0.045

0.05 0.06 0.07 0.08 0.09

81 .32 85.93 89.93 93.49 96.69

0.0010299 0.0010331 0.0010359 0.0010385 0.0010409

3.2401 2.7318 2.3649 2.0872 1.8695

340.48 359.84 376.68 391.64 405.13

2645.2 2652.9 2659.4 2665.2 2670.3

1.0910 1.1 452 1.1 919 1.2328 1.2694

7.5930 7.5311 7.4790 7.4339 7.3942

0.05 0.06 0.07 0.08 0.09

0.10 0.12 0.14 0.16 0.18

99.61 104.78 109.29 113.30 116.91

0.0010431 0.0010473 0.0010510 0.0010544 0.0010576

1.6940 1.4284 1.2366 1.0914 0.97753

417.44 439.30 458.37 475.34 490.67

2674.9 2683.1 2690.0 2696.0 2701.4

1.3026 1.3608 1.4109 1.4549 1.4944

7.3588 7.2976 7.2460 7.2014 7.1620

0.10 0.12 0.14 0.16 0.18

0.20 0.25 0.30 0.35 0.40

120.21 127.41 133.53 138.86 143.61

0.0010605 0.0010672 0.0010732 0.0010786 0.0010836

0.88574 0.71870 0.60579 0.52420 0.46239

504.68 535.35 561.46 584.31 604.72

2706.2 2716.5 2724.9 2732.0 2738.1

1.5301 1.6072 1.6718 1.7275 1.7766

7.1269 7.0524 6.9916 6.9401 6.8954

0.20 0.25 0.30 0.35 0.40

0.45 0.50 0.55 0.60 0.65

147.91 151.84 155.46 158.83 161.99

0.0010882 0.0010926 0.0010967 0.0011006 0.0011044

0.41390 0.37480 0.34259 0.31558 0.29258

623.22 640.19 655.88 670.50 684.22

2743.4 2748.1 2752.3 2756.1 2759.6

1.8206 1.8606 1.8972 1.931 1 1.9626

6.8560 6.8206 6.7885 6.7592 6.7321

0.45 0.50 0.55 0.60 0.65

0.70 0.80 0.90 1.00 1.10

164.95 170.41 175.36 179.89 184.07

0.0011080 0.0011 148 0.0011212 0.0011272 0.0011 330

0.27276 0.24033 0.21487 0.19435 0. 17744

697.14 721.02 742.72 762.68 781.20

2762.7 2768.3 2773.0 2777.1 2780.7

1.9921 20460 2 0944 2. 1384 2.1789

6.7070 6.6615 6.6212 6.5850 6.5520

0.70 0.80 0.90 1.00 1.10

VL

Vv

23.734 20.525 18.099 16.200 14.671

MPa

Steam Tables  331 Table 5. Properties of Saturated Water and Steam (Pressure) Press.

Temp.

Volume, m3j kg

MPa

oc

1.2 1.3 1.4 1.5 1.6

187.96 191.61 195.05 198.30 201.38

0.001 139 0.001144 0.001 149 0.001 154 0.001159

1.8 2.0

2.2

2.4 2.6

207.12 212.38 217.26 221,80 226 05

2.8 3.0 3.2 3.4 3.6

Enthalpy, kJjkg

Entropy, kJ/(kg ·K)

Press.

hl

hv

St_

Sv

MPa

0.16325 0.15117 0.1 4077 0.13170 0.12373

798.50 814.76 830.13 844.72 858.61

2783.8 2786.5 2788.9 2791.0 2792.9

2.2163 2.2512 2.2839 2.3147 2.3438

6.5217 6.4936 6.4675 6.4431 6.4200

1.2 1.3 1.4 1.5 1.6

0.0011 68 0.001177 0.001185 0.001193 0.001201

0.1 1036 0.099581 0.090695 0.083242 0.076897

884.61 908.62 930.98 951.95 971.74

2796.0 2798.4 2800.2 2801.5 2802.5

2.3978 2.4470 2.4924 2.5344 2.5738

6.3776 6.3392 6.3040 6.2714 6.241 1

1.8 2.0 2.2 2.4 2.6

230 06 233.86 237.46 240.90 244.19

0.001209 0.001217 0.001224 0.001231 0.001239

0.071428 0.066664 0.062475 0.058761 0.055446

990.50 1008.4 1025.5 1041.8 1057.6

2803.0 2803.3 2803.2 2803.0 2802.5

2.6107 2.6456 2.6787 2.7102 2.7403

6.2126 6.1 858 6.1604 6.1362 6.1131

2.8 3.0 3.2 3.4 3.6

3.8 4.0 4.2 4.4 4.6

247.33 250.36 253.27 256.07 258.78

0.001246 0.001253 0.001259 0.001266 0.001273

0.052468 0.049777 0.047333 0.045103 0.043060

1072.8 1087.4 1101.6 1115.4 1128.8

2801 .8 2800.9 2799.9 2798.7 2797.3

2.7690 2.7967 2.8232 2.8488 2.8736

6.0910 6.0697 60492 6.0294 6 0103

3.8 4.0 4.2 4.4 4.6

4.8 5.0 5.5 6.0 6.5

261.40 263.94 269.97 275.59 280.86

0.001280 0.001286 0.001303 0.001319 0.001336

0.041181 0.039446 0.035642 0.032449 0.029728

1141.8 1154.5 1184.9 1213.7 1241.2

2795.8 2794.2 2789.7 2784.6 2778.8

2.8975 2.9207 2.9759 3.0274 3.0760

5.9917 5.9737 5.9307 5.8901 . 5.8515

4.8 5.0 5.5 6.0 6.5

7.0 7.5 8.0 8.5 9.0

285.83 290.54 295.01 299.27 303.35

0.001352 0.001368 0.001385 0.001401 0.001418

0.027380 0.025331 0.023528 0.021926 0.020493

1267.4 1292.7 1317.1 1340.7 1363.7

2772.6 2765.8 2758.6 2751 .0 2742.9

3.1220 3.1658 3.2077 3.2478 3.2866

5.8146 5.7792 5.7448 5.7115 5.6790

7.0 7.5 8.0 8.5 9.0

9.5 10.0 11.0 12.0 13.0

307.25 311.00 318.08 324.68 330.86

0.001435 0.001453 0.001489 0.001526 0.001566

0.019203 0.018034 0.015994 0.014269 0.012785

1386.0 1407.9 1450.3 1491.3 1531.4

2734.4 2725.5 2706.4 2685.6 2662.9

3.3240 3.3603 3.4300 3.4965 3.5606

5.6472 5.6159 5.5545 5.4941 5.4339

9.5 10.0 11.0 12.0 13.0

14.0 15.0 16.0 17.0 18.0

336.67 342.16 347.36 352.29 356.99

0.001610 0.001657 0.001710 0.001769 0.001839

0.011489 0.010340 0.009308 0.008369 0.007499

1570.9 1610.2 1649.7 1690.0 1732.0

2638.1 2610.9 2580.8 2547.4 2509.5

3.6230 3.6844 3.7457 3.8077 3.8717

5.3730 5.3108 5.2463 5.1785 5.1055

14.0 15.0 16.0 17.0 18.0

19.0 20.0 21.0 22.0

361.47 365.75 369.83 373.71 373.946

0.001925 0.002039 0.002212 0.002750 0.00311

0.006673 0.005858 0.004988 0.003577 0.00311

1776.9 1827.1 1889.4 2021 .9 2087.5

2465.4 2411.4 2337.5 2164.2 2087.5

3.9396 4.0154 4.1093 4.3109 4.4120

5.0246 4.9299 4.8062 4.5308 4.4120

19.0 20.0 21.0 22.0

Pc

Pc=22 064 MPa

VL

Vv

Pc

332  Appendix B Table 6. Pressure MPa (Sat. T)

0.005 (32.88)

Su~erheated

Steam - SI Units

Temperature-Degrees Celsius 50 100 150 200

250

300

350

400

450

500

550

600

700

29.782 2593.4 8.4976

34.419 2688.0 8.7700

39.043 2783.4 9.0097

43.663 2879.8 9.2251

48.281 2977.6 9.4216

52.898 3076.9 9.6027

57.515 3177.6 9.7713

62.131 3280.0 9.9293

66.747 3384.0 10.078

71.363 3489.7 10.220

75.979 3597.1 10.354

80.594 3706.3 10.483

89.826 3929.9 10.725

V 14.867 h 2592.0 s 8.1741

17.197 2687.4 8.4488

19.514 2783.0 8.6892

21.826 2879.6 8.9048

24.136 2977.4 9.1014

26446 3076.7 9.2827

28.755 3177.5 9.4513

31.064 3279.9 9.6093

33.372 3384.0 9.7584

35.680 3489.7 9.8997

37.988 3597.1 10.034

40.296 3706.3 10.163

44.912 3929.9 10.405

V

8.5857 2686.2 8.1262

9.7488 2782.3 8.3680

10.907 2879.1 8.5842

12.064 2977.1 8.7811

13.220 3076.5 8.9624

14.375 3177.4 9.1311

15.530 3279.8 9.2892

16.684 3383.8 9.4383

17.839 3489.6 9.5797

18.993 3597.0 9.7143

20.147 3706.2 9.8431

22.455 3929.8 10.086

3.4188 2682.4 7.6952

3.8899 2780.2 7.9412

4.3563 2877.8 8.1591

4.8207 2976.2 8.3568

5.2841 3075.8 8.5386

5.7470 3176.8 8.7076

6.2095 3279.3 8.8658

6.6718 3383.5 9.0150

7.1339 3489.2 9.1565

7.5959 3596.7 9.2912

8.0578 3706.0 9.4200

8.9814 3929.7 9.6625

1.6960 2675.8 7.3610

1.9367 2776.6 7.6147

2.1725 2875.5 7.8356

2.4062 2974.5 8.0346

2.6389 3074.5 8.2171

2.8710 3175.8 8.3865

3.1027 3278.5 8.5451

3.3342 3382.8 8.6945

3.5656 3488.7 8.8361

3.7968 3596.3 8.9709

4.0279 3705.6 9.0998

4.4900 3929.4 9.3424

0.15 V (111.35} h

1.2856 2772.9 7.4207

1.4445 2873.1 7.6447

1.6013 2972.9 7.8451

1.7571 3073.3 8.0284

1.9123 3174.9 8.1983

2.0671 3277.8 8.3571

2.2217 3382.2 8.5067

2.3762 3488.2 8.6484

2.5305 3595.8 8.7833

2.6847 3705.2 8.9123

2.9929 3929.1 9.1550

0.20 V (120.21} h

0.9599 2769.1 7.2809

1.0805 2870.8 7.5081

1.1989 2971 .3 7.7100

1.3162 3072.1 7.8940

1.4330 3173.9 8.0643

1.5493 3277.0 8.2235

1.6655 3381.5 8.3733

1.7814 3487.6 8.5151

1.8973 3595.4 8.6501

2.0130 3704.8 8.7792

2.2444 3928.8 9.0220

0.25 V (127.41) h

0.7644 2765.2 7.1707

0.8621 2868.4 7.4013

0.9574 2969.6 7.6046

1.0517 3070.8 7.7895

1.1454 3172.9 7.9602

1.2387 3276.2 8.1196

1.3317 3380.9 8.2696

1.4246 3487.1 8.4116

1.5174 3594.9 8.5467

1.6101 3704.4 8.6759

1.7952 3928.5 8.9188

0.30 V (133.53) h

0.6340 2761.2 7.D791

0.7164 2866.0 7.3132

0.7965 2967.9 7.5181

0.8753 3069.6 7.7037

0.9536 3172.0 7.8749

1.0315 3275.4 8.0346

1.1092 1.1867 3380.2 3486.6 8.1848 8.3269

1.2641 3594.5 8.4622

1.3414 3704.0 8.5914

1.4958 3928.2 8.8344

0.35 V (138.86) h

0.5408 2757.1 7.0002

0.6124 2863.5 7.2381

0.6815 2966.3 7.4445

0.7494 3068.4 7.6310

0.8167 3171.0 7.8026

0.8836 3274.6 7.9626

0.9503 3379.6 8.1130

1.0168 3486.0 8.2553

1.0832 3594.0 8.3906

1.1495 1.2819 3703.6 3927.9 8.5199 8.7630

0.40 V (143.61) h

0.4709 2752.8 6.9305

0.5343 2861.0 7.1724

0.5952 2964.6 7.3805

0.6549 3067.1 7.5677

0.7139 3170.0 7.7398

0.7726 3273.9 7.9001

0.8311 3379.0 8.0507

0.8894 3485.5 8.1931

0.9475 3593.6 8.3286

1.0056 3703.2 8.4579

1.1215 3927.6 8.7012

0.45 V (147.91) h

0.4164 2748.3 6.8677

0.4736 2858.5 7.1139

0.5281 2962.8 7.3237

0.5814 3065.9 7.5117

0.6341 3169.0 7.6843

0.6863 3273.1 7.8449

0.7384 3378.3 7.9957

0.7902 3484.9 8.1383

0.8420 3593.1 8.2738

0.8936 3702.8 8.4032

0.9968 3927.3 8.6466

0.50 V (151.84} h

0.4250 2855.9 7.0611

0.4744 2961.1 7.2726

0.5226 3064.6 7.4614

0.5701 3168.1 7.6345

0.6173 3272.3 7.7954

0.6642 3377.7 7.9464

0.7109 3484.4 8.0891

0.7576 3592.6 8.2247

0.8041 3702.5 8.3543

0.8970 3927.0 8.5977

0.55 V (155.46) h

0.3853 2853.3 7.0128

0.4305 2959.4 7.2261

0.4745 3063.3 7.4158

0.5178 3167.1 7.5894

0.5608 3271.5 7.7505

0.6035 3377.0 7.9017

0.6461 3483.9 8.0446

0.6885 3592.2 8.1803

0.7308 3702.1 8.3099

0.8153 3926.8 8.5535

0.60 V (158.83) h

0.3521 2850.7 6.9684

0.3939 2957.7 7.1834

0.4344 3062.1 7.3740

0.4743 3166.1 7.5480

0.5137 3270.7 7.7095

0.5530 3376.4 7.8609

0.5920 3483.3 8.0039

0.6309 3591.7 8.1398

0.6698 3701.7 8.2694

0.7473 3926.5 8.5131

0.01 (45.81)

V

h

s

0.02 (60.06}

h

0.05 (8132)

h

0.10 (99.61)

h

V=

V

V

specific volume, m3/kg

h = enthalpy, kJ/kg

s = entropy, kJ/(kg ·K)

Steam Tables  333 Table 6 {continued}. Pressure MPa (Sat. T)

Su~erheated

Temperature-Degrees Celsius 200 250 300 350

Steam - SI Units

400

450

650

700

750

800

0.6181 0.6539 3701.3 3812.9 8.2321 8.3564

0.6897 3926.2 8.4759

0.7254 4041.1 8.5911

0.7611 4157.7 8.7024

0.4735 0.5070 0.5405 3375.1 3482.3 3590.8 7.7884 7.9317 80678

0.5738 3700.9 8.1976

0.6071 3812.6 8.3220

0.6403 3925.9 8.4415

0.6735 4040.8 8.5567

0.7067 4157.5 8.6680

0.4417 0.4731 0.5043 3374.4 3481.7 3590.4 7.7559 7.8994 8.0355

0.5354 3700.5 8.1654

0.5665 3812.2 8.2898

0.5975 3925.6 8.4094

0.6285 4040.6 8.5246

0.6595 4157.3 8.6360

0.4139 3373.8 7.7255

0.4433 3481.2 7.8690

0.4726 3589.9 80053

0.5019 0.5310 0.5601 3700.1 3811 .9 3925.3 8.1353 8.2598 8.3794

0.5892 4040.3 8.4947

0.6182 4157.1 8.6060

0.3145 3160.2 7.3538

0.341 1 0.3675 3266.0 3372.5 7.5172 7.6698

0.3938 3480.1 7.8136

0.4199 0.4459 0.4718 3589.0 3699.3 3811.2 7.9501 8.0803 8.2049

0.4977 3924.7 8.3246

0.5236 0.5494 4039.8 4156.6 8.4399 8.5513

0.2825 3158.2 7.3028

0.3066 3264.4 7.4668

0.3304 3371.2 7.6198

0.3541 0.3777 0.4011 0.4245 3479.0 3588.1 3698.6 3810.5 7.7640 7.9007 80309 8.1557

0.4478 3924.1 8.2755

0.4711 4039.3 8.3909

0.4944 4156.2 8.5024

0.2339 0.2563 3049.1 3156.2 7.0773 7.2564

0.2783 3262.8 7.4210

0.3001 0.3217 33699 34779 7.5745 7.7189

0.3431 0.3645 3587.2 3697.8 7.8558 7.9863

0.3858 0.4070 0.4282 3809.9 3923.5 4038.8 8.1 111 8.2310 8.3465

0.4494 4155.7 8.4580

0.2345 3154.1 7.2138

0.2548 3261.2 7.3791

0.2748 3368.6 7.5330

0.2946 3476.8 7.6777

0.3143 3586.2 7.8148

0.3339 3697.0 7.9454

0.3535 3809.2 8.0704

0.3730 0.3924 3922.9 4038.3 8.1904 8.3059

0.411 8 4155.2 8.4175

1.3 V 0.1552 0.1769 (191.61) h 2809.6 2931.8 s 6.5430 6.7888

0.1969 0.2161 3043.7 3152.1 6.9931 7.1745

0.2349 3259.6 7.3404

0.2534 3367.3 7.4947

0.2718 0.2900 3475.7 3585.3 7.6397 7.7770

0.3081 3696.2 7.9078

0.3262 3808.5 8.0329

0.3442 3922.4 8.1530

0.3621 4037.8 8.2686

0.3801 4154.8 8.3803

1.4 V 0.1430 (195.05) h 2803.0 s 6.4975

0.1635 2927.9 6.7488

0.1823 3041.0 6.9553

0.3028 0.3195 3807.8 3921.8 7.9981 8.1183

0.3362 4037.2 8.2340

0.3529 4154.3 8.3457

1.5 V 0.1324 (198.30) h 2796.0 s 6.4537

0.1520 0.1697 2924.0 3038.3 6.71 11 6.9199

0.1866 3148.0 7.1035

0.2825 3807.2 7.9657

0.2981 3921.2 8.0860

0.3137 0.3293 4036.7 4153.9 8.2018 8.3135

1.6 V (201.38) h

0.1419 2919.9 6.6754

0.1746 0.1901 0.2053 3146.0 3254.7 3363.3 7.0713 7.2392 7.3948

0.2203 3472.5 7.5407

0.2352 3582.6 7.6787

0.2500 0.2647 3693.9 3806.5 7.8099 7.9354

0.2794 3920.6 8.0557

0.2940 4036.2 8.1716

0.3087 4153.4 8.2834

1.7 V (204.31) h

0.1330 0.1489 29159 3032.7 6.6413 6.8548

0.1640 31439 7.0408

0.1930 3362.0 7.3654

0.2072 3471.4 7.5117

0.2212 3581.6 7.6499

0.2352 0.2491 3693.1 3805.8 7.7813 7.9068

0.2629 3920.0 8.0273

0.2767 4035.7 8.1432

0.2904 4153.0 8.2551

1.8 V (207.12) h

0.1250 2911.7 6.6087

0.1402 0.1546 0.1685 0.1821 3029.9 3141.8 3251.5 3360.7 6.8247 7.01 19 7.1812 7.3377

0.1955 3470.3 7.4842

0.2088 3580.7 7.6226

0.2220 3692.3 7.7542

0.2351 0.2482 3805.1 3919.4 7.8799 8.0004

0.2612 0.2743 4035.2 4152.5 8.1164 8.2284

2.0 V (212.38) h

0.11 15 0.1255 2903.2 3024.3 6.5474 6.7685

0.1635 0.1757 3358.1 3468.1 7.2863 7.4335

0.1877 3578.9 7.5723

0.1996 3690.7 7.7042

0.21 15 0.2233 3803.8 3918.2 7.8301 7.9509

0.2350 4034.2 8.0670

0.65 V 0.3241 0.3629 (161.99) h 2848.0 29559 s 6.9270 7.1439

0.4005 3060.8 7.3354

0.4374 0.4739 3165.1 3269.9 7.5099 7.6717

0.5102 3375.7 7.8233

0.70 V 0.3000 (164.95) h 2845.3 s 6.8884

0.3364 2954.1 7.1071

0.3714 3059.5 7.2995

0.4058 0.4398 3164.1 3269.1 7.4745 7.6366

0.75 V 0.2791 0.3133 (167.76) h 2842.5 ~952.3 s 6.8520 7.0727

0.3462 3058.2 7.2660

0.3784 0.4102 3163.1 3268.4 7.4415 7.6039

0.80 V 0.2609 (170.41) h 2839.8 s 6.8176

0.2932 2950.5 7.0403

0.3242 0.3544 0.3843 3056.9 3162.2 3267.6 7.2345 7.4106 7.5733

0.90 V 0.2304 (175.36) h 2834.1 s 6.7538

0.2596 2946.9 6.9806

0.2874 3054.3 7.1768

1.0 V 0.2060 (179.89) h 2828.3 s 6.6955

0.2327 2943.2 6.9266

0.2580 3051.7 7.1247

1.1 V 0.1860 (184.07) h 2822.3 s 6.6414

0.2107 2939.5 6.8772

1.2 V 0.1693 (187.96) h 2816.1 s 6.5908

0.1924 0.2139 2935.7 3046.4 6.8314 7.0336

V= specific

volume, m3Jkg

0.1587 3035.5 6.8865

500

0.2003 0.2178 0.2351 0.2522 3150.1 3258.0 3366.0 3474.7 7.1378 7.3044 7.4591 7.6045

0.1386 3137.6 6.9582

h = enthalpy, kJ/kg

0.2030 3256.4 7.2708

0.1786 3253.1 7.2094

0.1512 3248.2 7.1290

0.2192 3364.7 7.4259

550

0.5463 0.5822 3482.8 3591.3 7.9665 8.1024

600

0.2691 0.2860 3584.4 3695.4 7.7420 7.8729

0.2352 0.2510 0.2668 3473.6 3583.5 3694.6 7.5716 7.7093 7.8404

s =entropy, kJ/(kg·K)

0.2467 4151.6 8.1791

334  Appendix B Table 6 (continued}. Pressure MPa (Sat T)

Su~erheated

Temperature-Degrees Celsius 225 250 300 350

Steam - SI Units

400

450

500

600

650

700

750

800

2.2 V 0.0931 (217.26) h 2824.5 s 6.3531

0.1004 2894.5 6.4903

0.1134 3018.5 6.7168

0.1255 3133.4 6.9091

0.1371 3244.9 7.0813

0.1484 3355.4 7.2396

0.1595 3465.9 7.3873

0.1704 0.1813 3577.0 3689.1 7.5266 7.6588

0.1921 3802.4 7.7850

0.2028 3917.1 7.9059

0.2136 4033.1 8.0222

0.2242 4150.7 8.1344

2.4 V 0.0842 (221.80) h 2812.1 s 6.2926

0.0911 2885.5 6.4365

0.1034 3012.6 6.6688

0.1146 3129.1 6.8638

0.1253 3241.6 7.0375

0.1357 3352.7 7.1967

0.1459 3463.7 7.3450

0.1560 3575.2 7.4848

0.1660 3687.6 7.6173

0.1760 3801.1 7.7437

0.1858 3915.9 7.8648

0.1957 4032.1 7.9813

0.2055 4149.8 8.0936

2.6 V (226.05) h

0.0833 2876.2 6.3854

0.0948 3006.6 6.6238

0.1053 3124.8 6.8216

0.1153 3238.3 6.9968

0.1250 3350.0 7.1570

0.1345 3461.5 7.3060

0.1439 3573.3 7.4461

0.1531 3686.0 7.5790

0.1623 3799.7 7.7056

0.1 714 3914.7 7.8269

0.1805 4031.1 7.9435

0.1896 4148.9 8.0559

2.8 V (230.06) h

0.0765 2866.5 6.3365

0.0875 3000.5 6.5814

0.0974 3120.5 6.7821

0.1068 3234.9 6.9589

0.1 158 3347.4 7.1200

0.1247 3459.3 7.2696

0.1334 3571.5 7.4102

0.1420 3684.4 7.5434

0.1506 3798.4 7.6703

0.1591 3913.5 7.7918

0.1676 4030.0 7.9085

0.1760 4148.0 8.0210

V 3.0 (233.86) h

0.0706 2856.5 6.2893

0.0812 2994.3 6.5412

0.0906 3116.1 6.7449

0.0994 3231 .6 6.9233

0.1079 3344.7 7.0853

0.1162 3457.0 7.2356

0.1244 3569.6 7.3767

0.1324 3682.8 7.5102

0.1405 3797.0 7.6373

0.1484 3912.3 7.7590

0.1563 0.1642 4029.0 4147.0 7.8759 7.9885

3.2 V (237.46) h

0.0655 2846.2 6.2434

0.0756 2988.0 6.5029

0.0845 3111.6 6.7097

0.0929 3228.2 6.8897

0.1009 3341.9 7.0527

0.1088 3454.8 7.2036

0.1165 3567.7 7.3451

0.1240 3681.2 7.4790

0.1316 3795.6 7.6064

0.1390 0.1465 391 1.2 4028.0 7.7283 7.8453

0.1539 4146.1 7.9581

3.4 V (240.90) h

0.0609 2835.3 6.1986

0.0707 2981.6 6.4662

0.0792 3107.1 6.6762

0.0872 3224.8 6.8579

0.0948 3339.2 7.0219

0.1022 3452.6 7.1735

0.1095 3565.9 7.3154

0.11 66 3679.6 7.4496

0.1237 3794.3 7.5773

0.1308 3910.0 7.6993

0.1378 4026.9 7.8165

0.1 448 4145.2 7.9294

3.6 V (244.19) h

0.0568 2824.0 6.1545

0.0663 2975.1 6.4309

0.0745 3102.6 6.6443

0.0821 3221.3 6.8276

0.0893 3336.5 6.9927

0.0964 3450.3 7.1449

0.1033 3564.0 7.2873

0.1101 0.1168 3678.0 3792.9 7.4219 7.5498

0.1234 3908.8 7.6720

0.1301 4025.9 7.7893

0.1367 4144.3 7.9023

3.8 V (247.33) h

0.0531 2812.1 6.1107

0.0624 2968.4 6.3968

0.0703 3098.0 6.6137

0.0775 3217.9 6.7988

0.0844 3333.7 6.9649

0.091 1 3448.1 7.1178

0.0977 3562.1 7.2607

0.1042 3676.4 7.3955

0.1105 3791.5 7.5237

0.1169 0.1232 3907.6 4024.8 7.6461 7.7636

0.1294 4143.4 7.8767

4.0 V (250.36) h

0.0589 2961.7 6.3638

0.0665 3093.3 6.5843

0.0734 3214.4 6.7712

0.0800 3331 .0 6.9383

0.0864 3445.8 7.0919

0.0927 3560.2 7.2353

0.0989 3674.8 7.3704

0.1049 3790.2 7.4989

0.1110 3906.4 7.6215

0.1170 0.1229 4023.8 4142.5 7.7391 7.8523

4.5 V (257.44) h

0.0514 2944.1 6.2852

0.0584 3081.5 6.5153

0.0648 3205.6 6.7069

0.0708 3324.0 6.8767

0.0765 3440.2 7.0320

0.0821 0.0877 3555.5 3670.8 7.1765 7.3126

0.0931 3786.7 7.4416

0.0985 3903.4 7.5647

0.1038 4021.2 7.6827

0.1092 4140.2 7.7962

5.0 V (263.94) h

0.0453 2925.6 6.2109

0.0520 3069.3 6.4515

0.0578 3196.6 6.6481

D.0633 3317.0 6.8208

0.0686 3434.5 6.9778

0.0737 35508 7.1235

0.0787 3666.8 7.2604

0.0836 3783.3 7.3901

0.0885 3900.5 7.5137

0.0933 4018.6 7.6321

0.0982 4137.9 7.7459

5.5 V (269.97) h

0.0404 2906.2 6.1396

0.0467 3056.8 6.3919

0.0522 3187.5 6.5938

0.0572 3309.9 6.7693

0.0621 3428.7 6.9282

0.0668 3546.0 7.0751

0.0714 3662.8 7.2129

0.0759 3779.8 7.3432

0.0803 3897.5 7.4673

0.0848 4016.0 7.5861

0.0891 4135.6 7.7002

6.0 V (275.59) h

0.0362 2885.5 6.0702

0.0423 3043.9 6.3356

0.0474 3178.2 6.5431

0.0522 3302.8 6.7216

0.0567 3422.9 6.8824

0.0610 3541.2 7.0306

0.0653 3658.8 7.1692

0.0694 3776.4 7.3002

0.0735 3894.5 7.4248

0.0776 4013.4 7.5439

0.0816 4133.3 7.6583

6.5 V (280.86) h

0.0326 2863.5 6.0018

0.0385 3030.6 6.2819

0.0434 3168.7 6.4953

0.0479 3295.5 6.6771

0.0521 3417.1 6.8397

0.0561 3536.4 6.9892

0.0601 3654.7 7.1287

0.0640 3772.9 7.2603

0.0678 3891.5 7.3854

0.0716 4010.7 7.5050

0.0753 4131.0 7.6196

V= specific

volume, m3/kg

h = enthalpy, kJ/kg

s = entropy, kJ/(kg ·K)

550

Steam Tables  335 Table 6 {continued}. Pressure MP a (Sat. T)

Su~erheated

Temperature-Degrees Celsius 325 375 300 350

Steam - SI Units

400

450

500

550

600

650

700

750

800

7.0 V 0.0295 (285.83) h 2839.8 s 5.9335

0.0326 2935.5 6.0970

0.0353 3016.8 6.2303

0.0377 3090.4 6.3460

0.0400 3159.1 6.4501

0.0442 3288.2 6.6351

0.0482 3411.3 6.7997

0.0520 3531.5 6.9505

0.0557 3650.6 7.0909

0.0593 3769.4 7.2232

0.0628 3888.5 7.3488

0.0664 4008.1 7.4687

0.0698 4128.6 7.5837

7.5 V 0.0267 (290.54) h 2814.3 s 5.8644

0.0298 2917.4 60407

0.0325 3002.7 6.1805

0.0348 3078.8 6.3002

0.0370 3149.3 6.4070

0.0410 3280.7 6.5954

0.0448 3405.3 6.7620

0.0483 3526.7 6.9141

0.0518 3646.5 7.0555

0.0552 3765.9 7.1885

0.0586 3885.4 7.3145

0.0619 4005.5 7.4348

0.0651 4126.3 7.5501

8.0 V 0.0243 (295.01) h 2786.4 s 5.7935

0.0274 2898.3 5.9849

0.0300 2988.1 6.1319

0.0323 3066.9 6.2560

0.0343 3139.3 6.3657

0.0382 3273.2 6.5577

0.0418 3399.4 6.7264

0.0452 3521.8 6.8798

0.0485 3642.4 7.0221

0.0517 3762.4 7.1557

0.0548 3882.4 7.2823

0.0579 4002.9 7.4030

0.0610 4124.0 7.5186

8.5 V 0.0220 (299.27) h 2755.4 s 5.7193

0.0252 2878.3 5.9294

0.0278 2972.9 6.0845

0.0300 3054.7 6.2132

0.0320 3129.1 6.3259

0.0357 0.0391 3265.6 3393.4 6.5216 6.6925

0.0424 3516.9 6.8473

0.0455 3638.3 6.9905

0.0485 3758.9 7.1248

0.0515 3879.4 7.2519

0.0545 4000.2 7.3730

0.0574 4121.7 7.4889

9.0 V (303.35) h

0.0233 2857.0 5.8736

0.0258 2957.2 6.0378

0.0280 3042.2 6.1716

0.0300 3118.8 6.2875

0.0335 3257.9 6.4871

0.0368 3387.3 6.6601

0.0399 3511.9 6.8163

0.0429 3634.2 6.9605

0.0458 3755.4 7.0955

0.0486 3876.4 7.2231

0.0514 3997.6 7.3446

0.0541 4119.4 7.4608

9.5 V (307.25) h

0.0215 2834.4 5.8170

0.0240 2940.9 5.9917

0.0262 3029.4 6.1309

0.0281 3108.2 6.2502

0.0316 3250.2 6.4538

0.0347 3381.2 6.6291

0.0377 3506.9 6.7867

0.0405 3630.0 6.9319

0.0433 3751.9 7.0676

0.0460 3873.3 7.1957

0.0486 3994.9 7.3176

0.0512 4117.0 7.4341

10.0 V (311 00) h

0.0199 2810.2 5.7593

0.0224 2924.0 5.9458

0.0246 3016.2 6.0910

0.0264 3097.4 6.2139

0.0298 3242.3 6.4217

0.0328 3375.1 6.5993

0.0357 3501.9 6.7584

0.0384 3625.8 6.9045

0.0410 3748.3 7.0409

0.0436 3870.3 7.1696

0.0461 3992.3 7.2918

0.0486 4114.7 7.4086

11.0 V (318.08) h

0.0170 2755.6 5.6373

0.0196 2887.8 5.8541

0.0217 2988.7 6.0129

0.0235 3075.1 6.1438

0.0267 3226.2 6.3605

0.0296 3362.6 6.5430

0.0322 3491.9 6.7050

0.0347 3617.5 6.8531

0.0371 3741.2 6.9910

0.0395 3864.2 7.1207

0.0418 3987.0 7.2437

0.0441 4110.1 7.3612

12.0 V (324.68) h

0.0143 2688.4 5.4988

0.0172 2848.0 5.7607

0.0193 2959.5 5.9362

0.0211 3051.9 6.0762

0.0242 3209.8 6.3027

0.0268 3350.0 6.4902

0.0293 3481.7 6.6553

0.0317 3609.0 6.8055

0.0339 3734.1 6.9448

0.0361 3858.0 7.0756

0.0383 3981.6 7.1994

0.0404 4105.4 7.3175

13.0 V (33086) h

O.D151 2803.6 5.6635

0.0173 2928.3 5.8600

0.0190 3027.6 6.0104

0.0220 3192.9 6.2475

0.0245 3337.1 6.4404

0.0269 3471.4 6.6087

0.0291 3600.5 6.7610

0.0312 3726.9 6.9018

0.0332 3851.9 7.0336

0.0352 3976.3 7.1583

0.0372 4100.7 7.2771

14.0 V (336.67) h

0.0132 2752.9 5.5595

0.0155 2894.9 5.7832

0.0172 3002.2 5.9457

0.0201 3175.6 6.1945

0.0225 3324.1 6.3931

0.0248 3461.0 6.5648

0.0268 3591.9 6.7192

0.0288 3719.7 6.8615

0.0308 3845.7 6.9944

0.0326 3970.9 7.1200

0.0345 4096.0 7.2393

15.0 V (342.16) h

O.D115 2693.0 5.4435

0.0139 2858.9 5.7049

0.0157 2975.5 5.8817

0.0185 3157.8 6.1433

0.0208 3310.8 6.3479

0.0229 3450.5 6.5230

0.0249 3583.3 6.6797

0.0268 3712.4 6.8235

0.0286 3839.5 6.9576

0.0304 3965.6 7.0839

0.0321 4091.3 7.2039

16.0 V (347.36) h

0.0098 2617.0 5.3045

0.0125 2819.5 5.6238

0.0143 2947.5 5.8177

0.0170 3139.6 6.0935

0.0193 3297.3 6.3045

0.0214 3439.8 6.4832

0.0232 3574.6 6.6422

0.0250 3705.1 6.7876

0.0267 3833.3 6.9228

0.0284 3960.2 7.0499

0.0301 4086.6 7.1706

17.0 V (352.29) h

0.0112 2775.9 5.5384

0.0130 2917.8 5.7533

0.0158 3120.9 60449

0.0180 3283.6 6.2627

0.0199 3429.1 6.4451

0.0218 3565.9 6.6064

0.0235 3697.8 6.7534

0.0251 3827.0 6.8897

0.0267 3954.8 7.0178

0.0282 4081.9 7.1391

18.0 V (356.99) h

0.0100 2726.9 5.4465

0.0119 2886.3 5.6881

0.0147 3101.7 5.9973

0.0168 3269.7 6.2222

0.0187 3418.3 6.4085

0.0204 3557.0 6.5722

0.0221 3690.4 6.7208

0.0236 3820.7 6.8583

0.0251 3949.4 6.9872

0.0266 4077.2 7.1091

V= specific

volume, m3/kg

h =enthalpy, kJ/kg

s =entropy, kJ/(kg ·K)

336  Appendix B Table 6 (continued}. Pressure MPa (Sat. T)

Su~erheated

Steam- SI Units

Temperature- Degrees Celsius 450 425 400 375

475

500

550

600

650

700

V 20 (365.75) h

0.00768 0.00995 2602.4 2816.8 5.2272 5.5525

0.0115 2952.9 5.7510

0.0127 3061.5 5.9041

0.0138 3155.8 6.0322

0.0148 3241.2 6.1445

0.0166 3396.2 6.3390

0.0182 3539.2 6.5077

0.0197 0.0211 3675.6 3808.2 6.6596 6.7994

V 22 (373.71) h

0.00490 0.00826 0.00987 2354.0 2735.8 2897.8 4.8240 5.4050 5.6417

0.0111 3019.0 5.8124

0.0122 3121.0 5.9511

0.0131 3211.8 6.0704

00148 3373.8 6.2736

0.0163 3521.2 6.4475

0.0178 3660.6 6.6029

00108 0.0118 3084.8 3181.4 5.8720 5.9991

0.0134 3350.9 6.2116

O.Q106 3150.2 5.9298

24

V

h

26

V

h

28

V

h

30

V

h

35

V

h

40

45

50

60

70

V

h V

h V

h V

h V

h

80

90

V

h V

h

100

V

h

750

800

0.0225 0.0239 3938.5 4067.7 6.9301 7.0534 0.0204 3927.6 6.8776

0.0216 4058.2 7.0022

0.0148 0.0161 3502.9 3645.6 6.3910 6.5499

0.0174 00186 3782.8 3916.7 6.6946 6.8289

00197 4048.8 6.9549

0.0121 3327.6 6.1523

0.0135 3484.4 6.3374

0.0148 3630.4 6.5000

0.0160 3770.0 6.6473

0.0171 3905.8 6.7833

0.0182 4039.3 6.9107

0.00185 0.00385 0.00625 0.00762 0.00867 0.00957 0.0111 1809.1 2334.4 2695.8 2875.1 3007.7 31 17.9 3303.9 3.9635 4.7552 5.2841 5.5367 5.7170 5.8621 6.0953

0.0124 3465.7 6.2863

0.0136 0.0147 0.0158 0.0168 3615.1 3757.1 3894.8 4029.7 6.4527 6.6026 6.7405 6.8693

0.00206 0.00673 0.00850 0.00977 1872.5 2637.4 2837.4 2974.0 4.0731 5.2366 5.5289 5.7212

0.00192 0.00529 0.00731 0.00862 0.00967 1832.8 2510.6 2770.6 2926.1 3047.0 4.0059 5.0304 5.4106 5.6296 5.7942

0.0191 3795.5 6.7451

00102 0.0114 0.0126 3279.8 3446.9 3599.7 6.0403 6.2374 6.4077

0.0137 3744.2 6.5602

0.0147 0.0156 3883.8 4020.2 6.7000 6.8303

0.00170 0.00211 0.00344 0.00496 0.00606 0.00693 0.00835 0.00952 0.0106 1762.5 1988.4 2373.5 2671 .0 2857.3 2998.0 3218.1 3399.0 3560.9 3.8725 4.2140 4.7752 5.1945 5.4480 5.6331 5.9093 6.1229 6.3032

0.0115 3711.9 6.4625

0.0124 3856.3 6.6072

0.0133 3996.5 6.7411

0.00164 0.00191 0.00254 0.00369 0.00476 0.00562 0.00699 0.00809 0.00905 0.00993 1742.7 1931.1 2198.6 2511 .8 2740.1 2906.7 3154.6 3350.4 3521.8 3679.4 3.8290 4.1141 4.5037 4.9447 5.2555 5.4746 5.7859 6.0170 6.2079 6.3743

0.0107 3828.8 6.5239

0.0115 3972.8 6.6614

0.00160 0.00180 0.00219 0.00292 0.00382 0.00463 0.00594 0.00698 0.00788 0.00870 0.00945 1728.0 1897.6 2110.8 2377.3 2623.4 2813.4 3090.2 3301.5 3482.5 3647.0 3801.3 3.7939 4.0505 4.3612 4.7362 5.0710 5.3209 5.6685 5.9179 6.1197 6.2932 6.4479

0.0102 3949.3 6.5891

0.00179 0.00280 0.00530 0.00674 0.00780 0.00869 1792.0 2152.4 2611 .9 2820.9 2966.7 3084.8 3.9314 4.4750 5.1473 5.4419 5.6402 5.7956

0.00156 0.00173 0.00201 0.00249 0.00317 0.00389 0.00512 0.00611 0.00696 0.00772 0.00842 0.00907 1716.6 1874.3 2060.2 2284.4 2520.0 2722.5 3025.7 3252.6 3443.5 3614.8 3774.1 3926.0 3.7642 4.0028 4.2738 4.5892 4.9096 5.1759 5.5566 5.8245 6.0372 6.2180 6.3777 6.5226 0.00150 0.00163 0.00182 0.00208 0.00247 0.00295 0.00395 0.00483 0.00559 0.00627 0.00688 0.00746 1699.9 1843.1 2001.6 2179.8 2375.2 2570.4 2902.1 3157.0 3366.8 3551.4 3720.6 3880.2 3.7148 3.9316 4.1626 4.4134 4.6790 4.9356 5.3519 5.6528 5.8867 6.0815 6.2512 6.4034 0.00146 0.00157 0.00171 0.00189 0.00214 0.00246 0.00322 0.00397 0.00465 0.00525 0.00580 0.00632 1688.4 1822.9 1967.1 2123.4 2291.7 2466.2 2795.0 3067.5 3293.6 3490.5 3669.0 3835.8 3.6743 3.8778 4.0880 4.3080 4.5368 4.7662 5.1786 5.5003 5.7522 5.9600 6.1390 6.2982 0.00143 0.00152 0.00163 0.00177 0.00196 0.00219 0.00276 0.00338 0.00398 0.00452 0.00501 0.00548 1680.4 1808.8 1944.0 2087.6 2239.6 2397.6 2709.9 2988.1 3225.7 3432.9 3619.7 3793.3 3.6395 3.8339 4.0311 4.2331 4.4398 4.6474 5.0391 5.3674 5.6321 5.8509 6.0382 6.2039 0.00140 0.00148 0.00157 0.00169 0.00184 0.00201 0.00246 0.00297 0.00348 0.00397 0.00442 0.00484 1674.6 1798.6 1927.6 2062.7 2204.0 2350.3 2645.2 2920.8 3164.4 3379.5 3573.5 3753.0 3.6089 3.7965 3.9847 4.1747 4.3669 4.5593 4.9288 5.2540 5.5255 5.7526 5.9470 6.1184 0.00137 0.00144 0.00153 0.00163 0.00175 0.00189 0.00225 0.00267 0.00311 0.00355 0.00395 0.00434 1670.7 1791.1 1915.5 2044.5 2178.3 2316.2 2596.1 2865.1 31 106 3330.8 3530.7 3715.2 3.5815 3.7638 3.9452 4.1267 4.3086 4.4899 4.8407 5.1580 5.4316 5.6640 5.8644 6.0405

v= specific volume, m3fkg

h = enthalpy, kJ/kg

s = entropy, kJ/(kg ·K)

Appendix C Common Units and Unit Conversion Factors Table Conv-1. Conversion Factors for Pressure (Force/Area)

337

To obtain multiply by

at m

bar

psi a (lb,lin1 )

at m

1

1.013 25

14.695 95

1

14.503 77

6.894 757x 10'2

3.386 388x 10'2

bar psi a Ob,lin1 )

11(1.013 25) = 9.869 233xiO·'

6.804 596x 10'2

25.41760 in Hg (conventional) = 3.342 I05x 10'2

inHg (conventional)

mmHg (conventional)

ftH1 0 (conventional)

kPa

MPa

760

33.898 54

101.325

0.101 325

29.529 99

760/(1.013 25) = 750.0617

33.455 26

100

0.1

1

2.036 021

51.714 93

2.306 659

6.894 757

6.894 757x 10'3

4.911541xi0' 1

1

25.4

1.132 925

3.386 388

3.386 388x 10'3

1

4.460 334x 10'2

101.325/760

(0.101 325)/760

760/25.4 = 29.921 26

(1.013 25)1760 11760 mmHg (conventional) = 1.315 789x 10'3 = 1.333 224X 10'3

1.933 677x 10·2

ftH 20 (conventional)

2.949 980x 10'2

2. 989 067 x 1o·2

4.335 275x 10'1

8.826 711xl0·'

22.419 85

1

2.989 067

2.989 067x 10·3

kPa

1/101.325 = 9.869 233 X10'3

0.01

1.450 377x 10'1

2.952 999xi0'1

760/101.325 =7.500617

3.345 526X Jo·l

1

0.001

MPa

11(0.101 325) = 9.869 233

10

145.0377

295.2999

334.5526

1000

1

11(25.4) = 3.937 008xl0'

2

760/(0. 101 325) 3

= 7.500 6J7x 10

= 1.333 224x 10·' = 1.333 224x 10-4

338  Appendix C

Table Conv-2. Conversion Factors for Specific Volume (Volume/Mass) Toobtainmultiply by

f~/lbm in31Ibm

ft3/lbm

in31Ibm

1

1728

1/1728

= 5.787 037xl0-4 231/1728

1

US gal/lbm

1728/231

= 7_480 519 1/231

=4.329 004x 10'3

liter/kg (cm3/g)

mJ/kg

6.242 796x 101

6.242 796x 10'2

3.612 729x 10'2

3.612 729xl0'5

US gal/lbm

= 1.336 806xl0-l

231

1

8.345 404

8.345 404x 10'3

liter/kg (cm3/g)

1.601 846xl0-2

2.767 990x101

1.198 264x 10'1

1

0.001

mJ/kg

1.601846x101

2.767 990x104

1.198 264x102

1000

1

Table Conv-3. Conversion Factors for Specific Enthalpy and Specific Energy (Energy/Mass) To obtain - _.Btu/lb., multiply bY..,_ Bt. b ..

1

bp·bllb.,

kW·bllb.,

psial(lb.,/ft')

callg

kJikg

7.781 693 x 102

3.930 148x 10""

2.930 7ll x JO""

5.403 953

1/1.8

4.1868/1.8

= 5.555 556 x 10-1

= 2.326

7.139 264 x 10""

2.989067x J0·3

1.413 574 x 103

5.918 353 x J03

l/(1.98X106)

1/144

fHb,!lb.,

1.285 067 x w·'

1

= 5.050 505 x w·'

3.766 161 x J0·7

= 6.944 444 x w·'

bp·bllb.,

2.544 434 x lo'

1.98x106

1

7.456 999 x w-l

(1.98x10")/144

kW·bllb.,

3.412 142 x 103

2.655 224 x 106

1.341 022

1

1.843 905 x 10 4

1.895 634 x lo'

7.936 641 x i03

psial(lb.,/ft')

1.850 497 x w·l

144

= 7.272 727 x 10·5

1

1.028 054 x w-l

4.304 256 x 10·1

cal/g

1.8

1.400 705 x 103

7.074 266x iO""

5.275 279x JO""

9.727116

1

4.1868

kJ/kg

= 4.299 226 x w-l

1/2.326

3.345 526x lo"

1.689 659x 10""

1.259 979x 10""

2.323 282

= 2.388 459 x w-l

144/(1.98x106)

5.423 272 x

w-5

= 1.375X104

1/4.1868

1

Common Units and Unit Conversion Factors  339

fHb,!lb.,

To obtain- • Btu/(lb,.·•R) multiply by'

fNbr/(lb,.··R)

kWh/(lb,.·•R)

psia·ft'/(lb,.· 0 R)

barcm3 /(g·K)

call(g·K)

kJ/(kg·K)

BJ(Ib,.·•R)

1

7.781 693xt02

2.930 711 xl0 4

5.403 953

41.868

1

4.1868

ft-lbrl(lb,.·"R)

1.285 067 x 10'3

1

3.766 16IxJO·'

11144 ; 6. 944 444x 10'3

5.380 320 x 10'2

1.285 067x 10·3

5.380 320x 10' 3

kW·hl(lb,.·•R)

3.412 142 x103

2.655 224 x106

1

1.843 905 x104

1.428 595 x 105

3.412 142 x103

1.428 595 x 104

144

5.423 272 x 10'5

1

7.747 661

1.850 497 x 10' 1

7.747 66t x tO·'

psia·ft3 /(lb,. ·"R) 1.850 497x to·' barcm3/(g· K)

0.1/4.1868 ; 2.388 459x w-2

1.858 625 x t0 1

6.999 882x10-6

1.290 712 x to·'

1

0.1/4.1868 ; 2.388 459 x 10'2

0.1

cal/(g·K)

1

7.781 693 x 102

2.930 7II xiO"'

5.403 953

41.868

1

4.1868

kJ/(kg·K)

114.1868 ; 2.388 459x to·'

1.858 625xJ02

6.999 882 x 10·5

1.290 712

10

1/4.1868 ; 2.388 459 x 10' 1

1

340  Appendix C

Table Conv-4. Conversion Factors for Specific Entropy, Heat Capacity, and Gas Constant (Energyj(Mass•Temperature))

Appendix D

Common Symbols Greek Alphabet Aa

Alpha

Nv

Nu

BJ3

Beta

:::(

Xi

rv

Gamma

Oo

Omicron

M

Delta

nn

Pi

EE

Epsilon

Pp

Rho

z~

Zeta

Io~

Sigma

HI]

Eta

TT

Tau

08

Theta

Vu

Upsilon

IL

Iota

Phi

KK

Kappa

xx

Chi

1\A

Lambda

liJljJ

Psi

MIJ.

Mu

nw

Omega

341

Index

A absolute atmospheric pressure  90 absolute enthalpy  41 absolute pressure  40–41, 66, 90 absolute temperature  5–7, 43 absolute zero  6 acceleration due to gravity  97 adiabatic  187, 199 and isenthalpic throttling process 148 open flow system  196 process 149, process example  149 thermodynamic process  148 air conditioning systems closed  229 large chiller-based  228 types of  227 air filtration sock filters  230 alarms 248 algorithm 252 alternative representation of Carnot heat cycle, p versus n 185 ammonia 232 analog input module  254 analog inputs  252, 254 analog output module  254 analog outputs  252, 254 Annual cost of energy  243 automated closed loop control systems 248 automated HVAC  227, 248–249, 251–252, 254

automated HVAC system  248–249, 251 auxiliary contacts  252 B Bernoulli’s equation  195 binary cycle  189 biomass combustion systems  164 blower or pump motors  254 BMS, building management systems 249 boiler with superheating function 167 boiling point  76, 232 Boyles Law  160 brake horsepower  24, 29, 167, 168 Btu 10 Btu’s to kWh  10 Btu’s to tons  10 C Carnot and Rankine cycle comparison 188 Carnot cycle 186, 188 Carnot heat cycle, h versus s  185 Carnot heat cycle, T versus s  184 central controllers  248 chamber properties  196 change in entropy  186 chemical formulas  232 chilled water control valve  230 chilled water spray nozzles  229

343

344  Index chlorofluorocarbons (CFC)  232 closed system  176 coefficient of performance, or COP 242 cogeneration 190 cogeneration and combined cycles 189 combined cycle  191 combined heat and power cycle  190 common energy conversion factors 33 common power conversion factors 32 common units and unit conversion factors 337 comprehensive EMS  249 compressible liquid  88 compressor  237, 246 compressor pump  186 compressors, rotary, reciprocating, centrifugal, screw, scroll  229 concentrated solar power tower  164 condenser  167, 171, 187, 231 condenser cooling tower for refrigeration system, forced-air type 233 condenser pump  191 conditioned air flow  230 constant entropy  158, 186 process 161 constant force  13 constant torque  13 constant volume process  156 example 157 controlled events, synchronous  248 controllers 248 control logic  248 convectional heat transfer 46 conveyor speed  30 cooling tower  190, 231 condenser 191

fan 231 cooling water  190 COP, computation  242 cost reduction  249 coupled moment 167, 168 couple in an impulse turbine 168 CPU 251 central processing unit 249 memory 251 critical constants  199 critical point  86–87, 89, 195, 199 pressure 88 critical pressure  87 ratio 199 critical ratios  199 critical temperatures  232 crystallographic forces  133 D DDC, direct digital controller  249 deca therms to Btu’s  10 degrees of freedom of the gas molecules 147 densities of common materials  17 density 3 of ice  105, 130 of water  130 dew point (DP)  203, 204, 205, 207 temperature 209 digital or discrete inputs  252 digital or discrete outputs  252 digital value  254 direction of flow of gas  199 discrete input block  252 district heating  190 double interpolation approach  62 double interpolation method  61 dry bulb  203–205, 207, 211, 213 change 220 temperature 210

Index  345

dry vapor  241 compression cycle in refrigeration systems 242 compression process  241 dynamic frictional force  12 E economic justification  249 efficiency 33 of the motor  29 Einstein’s equation  11 elastic energy  2 electrical energy  2, 163, 243 electromagnetic energy  2 electromotive force  162 eliminators 230 EMS, energy management systems 249 energy  1, 5 conservation 16 content and cost comparison  122 efficiency 140 productivity improvement projects 249 stored in a spring  8 unit conversion  10–11 energy engineer  243 enthalpy  34, 38, 45, 171, 173, 197, 201, 203, 237 determination 51 difference 136 -entropy chart  47 -entropy diagram  47 of the steam  128 scale 211 entropy  5, 6, 41, 45, 97, 171, 173, 174 analysis 42 change in  118 increase in  41 ESCO/EPC programs  249

evaporator  236, 238, 241, 246 event logs  251 exit-ambient conditions  196 expansion 238 valve  235, 238 external work  12 F feed pump  171 feedwater heaters, multiple  189 final humidity ratio  212 final point wet bulb  218 fire-tube boiler design  158 first law of thermodynamics  93, 94 flow rate of heat  129 fluid dynamics  238 fluorocarbons (FC)  232 force 22 formula for single interpolation  60 formulas for conversion of temperatures 6 freezing point  575 frictional force  12 frictional head loss  197 fuel heat content  121 fusion 132 G gas 75 gas dynamics  195 gaseous phase  76 gases 5 high velocity  195 gas laws, ideal  27 gc 8 generator 34 efficiency 37 output 37 generator 191 geometric analyses  212

346  Index geothermal 189 electrical power generation process 165 fluid 191 power generation  163, 164 g/gc  2–3, 8 global warming potential  232 graphical depictions of the HVAC systems 248 graphs 248 gravitational acceleration  3 gravitational constant  97 gravitational energy  2 gravitational force  14 GWP 232 H heat cycle in a Carnot engine p versus n 183 heat cycle in a heat engine without superheat  165, 166 heat cycle in a Rankine engine with superheat 172 heat cycle in a Rankine Engine with superheat, h versus n 174 heat cycle in a Rankine engine with superheat, h versus s  192 heat cycle in a Rankine engine with superheat, p versus n 172 heat cycle in a Rankine Engine with superheat, T versus s  173 heat cycle in a Rankine Engine with superheat 166 heat energy  10, 164 heat engine cycle  153 heat engine energy flow diagram  161 heat engine process flow diagram 161 heat exchange  231 with the surroundings  91 heat exchanger  231

heat of evaporation value  315 heat of fusion  109, 132 for ice or water  109 of water  236 heat of vaporization  55, 56 heat recovery steam generator  191 heat sink  161 heat transfer  42 HFC-134a 232 highly superheated vapor  86 high pressure liquid phase  231 high pressure liquid state  235 high pressure refrigerant vapor  238 high velocity flow  195 case study  198 high velocity gas system  196 HMI system  251 horsepower rating of the conveyor motor 28 human machine interface or HMI system 251 humidity ratio  203, 205, 208, 213 initial 222 lines 207 HVAC control system architecture, automated  250, 253 HVAC graphics  251 HVAC system architecture diagram 251 HVAC systems, direct digital control 248 hydroflourocarbons 232 I ideal gas  86, 148 laws 21 ideal heat engine—Carnot cycle 159 IEEE, Institute of Electrical and Electronic Engineers  249

Index  347

impulse turbine  167 design principle  170 initial dew point  220 initial humidity ratio  213 injection well  190 input energy  243 input modules  251 input power  33 internal combustion gasoline engine 147 internal energy  9, 39, 55 change in  93 per unit mass  9 internal work  12 interpolation  141, 208 final step in double process  128 second step  128 single and double  124 inversion point  147, 158 irreversible process  161 isenthalpic 149 isenthalpic (constant enthalpy) process 158 isenthalpic process example  149 isentropic  177, 185, 199 compression 188 process 153 expansion of vapor  186 flow 201 process  148, 158, 159 isobar 151 isobaric  173, 176, 184 condenser process  150 evaporator process  150 process 151 example  151, 152 segments of an ideal cycle heat engine 152 stages 153 thermodynamic process  151 isochoric process  156

isoenthalpic  149, 177 isometric process  156 isotherm  51, 150 isothermal  42, 174, 184 compression of vapor  149 compression process  155 expansion of liquid  185 process  153, 154, 173, 174 J Joule’s constant  40 K Kelvin temperature conversion  7 Kevin-Planck statement  98 kinetic 196 energy  1, 4, 9, 26 change in  93 kmols 155 kWh to Btu’s  10 L latent heat  50, 76, 84, 222 calculation 132 of fusion  76 of sublimation  76 of vaporization  77, 111, 135 flow rate  135 for water  111 law of conservation of energy  7 laws of thermodynamics  91, 93 light energy  2 liquid 73 and vapor mixture  240 phase 88 refrigerant 240 -vapor mixture  150 -vapor phase  79 logic level HIGH or “1”  252 logic level, LOW  252 low pressure liquid state  235

348  Index M mass flow across  93 mass flow rate  23, 96, 128 analysis 104 of the system  96, 192 mass of dry air  222 mass of refrigerant  240 mathematical equations for work  13 mechanical energy  162 mechanical system, work  12 microprocessor 251 mixture of liquid and vapor phases 48 MMBtu 10 moisture condensed/removed, amount of  217 molar enthalpy  40 molar internal energy  10, 40 molar specific volume  41 molecular kinetic energy  6 molecular weight  40 of common gases  157 Mollier Diagram  47, 50, 201 and the Steam Tables  71 application of  51 monitoring terminal  251 N natural gas energy content  116, 138 natural gas transportation cost  116, 139 negative work  12, 14, 153 non-adiabatic  171, 173, 184, 186 process 140 isobaric 238 non-CFC 232 non-HFC refrigerants  232 nozzle 199 number of moles  154

O ODP, ozone depletion Potential  232, 234 open-flow 196 open system steady flow energy equation 96 organic working fluid  189 outdoor air dampers  230 output energy  33 output modules  251 output power  33 ozone depletion potential  232 ozone layer  232 P Pascal 4 PC’s 248 phase conversion processes  87 phases of a substance  75 phases of water  75 phase transformation of water  83 phase transformation process  84 PLC, programmable logic controller 248 PLC’s 248 polytropic processes  160 positive work  14 potential 196 energy  2, 8, 27 change in  93 stored in the spring  20 power  22, 28 and efficiency  31 demanded from the utility  243 flow diagram  34 generation system  91 practical Rankine cycle  164 pressure  4, 83, 171, 174, 235 energy  2, 9, 10, 156 -enthalpy diagram  244, 246, 256

Index  349

-enthalpy diagram, HFC-134a 256 -enthalpy graph  238 process flow in a rankin cycle with superheat 171 production performance data diagnostics 251 programmable logic controllers  248 psychrometric analysis  203, 210 psychrometric chart  203, 207 interpretation guide  213 method for reading  205 psychrometric conditions  203, 209 psychrometric point  207 psychrometric processes  209 psychrometric transition  212 process 210 psychrometry 203 pump 167 Q quality  64, 246 quasiequilibrium processes  160 R radiated heat transfer  42 radiation 42 RAM, random access memory  251 Rankine cycle  164, 175, 189 engines  163, 166 equations 177 heat engine  121, 164 with regeneration  189 with superheat and reheat  189 Rankine and Carnot cycles comparison 188 temperature versus entropy  188 Rankine temperature conversion  7 ratio of specific heats  199

reaction turbine  170 design and operating principle 170 nozzles 169 real gas  86, 147 real time data  251 points 248 refrigerant  228, 236, 238, 241 coils 238 compression 238 process 241 refrigerant compressor power  241 refrigerants 231 commonly used  208 high-pressure 235 properties of  233 vapor quality ratio  240 refrigeration cycles  227, 241 and HVAC systems  227 basic 237 pressure-enthalpy graph  238 process flow diagram  237 refrigeration systems  204, 228 capacity quantification in A/C tons 236 chiller vs. compressor  228 compressors 231 condenser 231 cooling capacity of  236 thermal expansion valve  235 refrigerator compressor  228 regression analysis based forecasts 248 rejected energy  190 rejected heat  190 relative humidity  203 relative humidity line  207, 213 relief dampers  230 reservoir 199 reservoir pressure  199

350  Index residence time  196 reversible adiabatic process  147 RH 210 ROM, read only memory  251 rotational kinetic energy  192 rotational system  13 rotor 163 RTD, resistance temperature devices  204, 254 S safety interlock  252 saturated liquid  79 enthalpy 53 line 182 phase 79 state 79 saturated steam tables  55 saturated vapor  76 curve 173 enthalpy 53 line 182 saturated water  173 and steam tables, by pressure  56, 58 and steam tables, by temperature  56, 58 saturation curve  207 saturation line  47, 177 saturation pressure  58, 77 saturation temperature  59, 63, 77 second law of thermodynamics  6, 97 SEER or seasonal energy efficiency ratio 243 sensible heat  76, 83, 210 calculation  131, 138 removed, amount of  215 ratio  203, 215 sensible rise in temperature  173

set-points 248 SFEE  94, 98, 102, 196 shock waves  201 SHR, sensible heat ratio  212 single and double interpolation  60 single line diagram  164 solid 75 solid phase  84 solids 5 sonic pressure  199 sound energy  2 specific enthalpy  39, 40, 205, 208 change in  161 lines 209 specific entropy  58 specific gas constant  154 specific heat  15, 16, 95 specific heat, cp, for select liquids and solids  103 specific heat of ice  108 specific heat of water  111, 134 specific internal energy  40 specific volume  3, 40, 57, 171, 173 lines 222 units for  4 specific weight  3 specific work  95, 154 speed of light  10 spring constant  9 spring expansion or contraction  14 stagnation properties  196 static frictional force  12 steady flow energy equation  94, 195 steady flow open system  92 steam engine  163 steam generation process  154 steam power generation system  35 steam superheating stage  192 steam tables  315 steam to wire energy conversion  35

Index  351

steam to wire power flow  35 steam turbine efficiency  35 subcooled  63, 150 subcooled liquid  77, 177 state  76, 177 subcooled region  84 subsonic speed  199 sulfur dioxide  232 superconductivity 6 supercritical fluid phase  88 superfluidity 6 superheated 241 realm 184 steam  47, 181, 184 steam enthalpy  38 steam generation  157 steam reservoir  200 steam tables  59 superheated vapor  80, 174 region 84 state 81 supply air dampers  230 supply air fan  229 system boundary  92 system return or supply fan  238 T temperature  4, 83 change in  15 rise in  21 versus entropy graph  175 thermal efficiency of the entire Rankine cycle  178 thermal energy  1, 196 thermal equilibrium  5 thermocouples  204, 254 thermodynamic equilibrium  160 thermodynamic processes  145 in a typical refrigeration cycle 150

flow 100 reversible 161 thermodynamic systems  91 closed  92, 94 open  91, 92, 94–96 with a turbine  95 isolated 92 major categories  91 type of  91 thermophysical properties  204 third law of thermodynamics  42 throat 196 throat area of cross section  197 throttling of the high pressure liquid refrigerant 238 throttling process  148, 158, 168, in a refrigeration system  148, 149 timers 248 time schedules  248 tonne of refrigeration  237 torque 1 total properties  196 total volume of the air in the system 222 transformation of heat energy to mechanical work  163 transition of heat, mass and work in thermodynamic systems  93 trend charts  251 trend logs  248 triple point  6, 87–89 vs. critical point  87 turbine  34, 168, 175, 193 U unit conversion  10, 28 units for energy  32 units for power  32 units for pressure  4

352  Index universal gas constant  154, 155 utilities engineers  248 V vapor-liquid mixture  240 vapor phase  88 vapor, saturated  240 vapor, slightly superheated  239 vapor superheating stage  164 variable force  13 variable torque  13 velocity 22 of gas  196 of steam  128 W washer system architecture, open air 230 weight density  3 wet bulb  203, 207, 211 temperature  204, 207, 209, 213 wet compression cycle  239 wet vapor compression cycle in refrigeration systems  240

wet vapor compression process  239 work  1, 8, 12 external and internal  12 flow across the boundary  93 mathematical equations for  13 negative  12, 14, 148 performed  33, 154 against friction  19, 25 by the surroundings  149 by the system  92 by the turbine shaft  95 in a thermodynamic system  15 positive 167 producing turbine stage  193 working fluid  163, 185, 228, 238 working principle of a binary  190 cycle system  164 working principle of a combined cycle system  191 working principle of Rankine cycle engine with superheat and reheat 189

About the Author

Mr. S. Bobby Rauf is the President, Chief Consultant and a Senior Instructor at Sem-Train, LLC. Mr. Rauf has over 25 years of experience in teaching undergraduate and post-graduate engineering, science, math, business administration and MBA courses, seminars, and workshops. Mr. Rauf earned his BSc in Electrical Engineering, with honors, from NC State University, USA. He earned his Executive MBA degree from Pfeiffer University, USA. He is a registered Professional Engineer in the States of Virginia, Wyoming, and North Carolina and is a Certified Energy Manager. He holds a patent in process controls technology. Mr. Rauf’s last full-time engineering employment, in the corporate world, was at PPG Industries, Inc. where he served as a Senior Staff Engineer. Mr. Rauf was inducted as “Legend in Energy” by AEE, in 2014. Mr. Rauf is a member of ASEE, American Society of Engineering Education, and has developed and instructed Professional Engineering and Fundamentals of Engineering Exam (NCEES) Prep Courses over the past twenty years. Mr. Rauf develops and instructs PDH (Professional Development Hour) and continuing education, engineering skill building seminars and courses. See testimonials from past attendees at www.sem-train.com. He is also an Adjunct Professor at Gardner-Webb University. Mr. Rauf has published multiple texts over the last ten years.

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